Schaum's Outline of College Algebra [Fourth ed.] 0071821813, 9780071821810

Table of contents :
Cover
Title Page
Copyright Page
Contents
Chapter 1 Fundamental Operations with Numbers
1.1 Four Operations
1.2 System of Real Numbers
1.3 Graphical Representation of Real Numbers
1.4 Properties of Addition and Multiplication of Real Numbers
1.5 Rules of Signs
1.6 Exponents and Powers
1.7 Operations with Fractions
Chapter 2 Fundamental Operations with Algebraic Expressions
2.1 Algebraic Expressions
2.2 Terms
2.3 Degree
2.4 Grouping
2.5 Computation with Algebraic Expressions
Chapter 3 Properties of Numbers
3.1 Sets of Numbers
3.2 Properties
3.3 Additional Properties
Chapter 4 Special Products
4.1 Special Products
4.2 Products Yielding Answers of the Form a[sup(n)]+b[sup(n)]
Chapter 5 Factoring
5.1 Factoring
5.2 Factorization Procedures
5.3 Greatest Common Factor
5.4 Least Common Multiple
Chapter 6 Fractions
6.1 Rational Algebraic Fractions
6.2 Operations with Algebraic Fractions
6.3 Complex Fractions
Chapter 7 Exponents
7.1 Positive Integral Exponent
7.2 Negative Integral Exponent
7.3 Roots
7.4 Rational Exponents
7.5 General Laws of Exponents
7.6 Scientific Notation
Chapter 8 Radicals
8.1 Radical Expressions
8.2 Laws for Radicals
8.3 Simplifying Radicals
8.4 Operations with Radicals
8.5 Rationalizing Binomial Denominators
Chapter 9 Simple Operations with Complex Numbers
9.1 Complex Numbers
9.2 Graphical Representation of Complex Numbers
9.3 Algebraic Operations with Complex Numbers
Chapter 10 Equations in General
10.1 Equations
10.2 Operations Used in Transforming Equations
10.3 Equivalent Equations
10.4 Formulas
10.5 Polynomial Equations
Chapter 11 Ratio, Proportion, and Variation
11.1 Ratio
11.2 Proportion
11.3 Variation
11.4 Unit Price
11.5 Best Buy
Chapter 12 Functions and Graphs
12.1 Variables
12.2 Relations
12.3 Functions
12.4 Function Notation
12.5 Rectangular Coordinate System
12.6 Function of Two Variables
12.7 Symmetry
12.8 Shifts
12.9 Scaling
12.10 Using a Graphing Calculator
Chapter 13 Linear Equations in One Variable
13.1 Linear Equations
13.2 Literal Equations
13.3 Word Problems
Chapter 14 Equations of Lines
14.1 Slope of a Line
14.2 Parallel and Perpendicular Lines
14.3 Slope–Intercept form of Equation of a Line
14.4 Slope–Point Form of Equation of a Line
14.5 Two-Point Form of Equation of a Line
14.6 Intercept Form of Equation of a Line
Chapter 15 Simultaneous Linear Equations
15.1 Systems of Two Linear Equations
15.2 Systems of Three Linear Equations
Chapter 16 Quadratic Equations in One Variable
16.1 Quadratic Equations
16.2 Methods of Solving Quadratic Equations
16.3 Sum and Product of the Roots
16.4 Nature of the Roots
16.5 Radical Equations
16.6 Quadratic-type Equations
Chapter 17 Conic Sections
17.1 General Quadratic Equations
17.2 Conic Sections
17.3 Circles
17.4 Parabolas
17.5 Ellipses
17.6 Hyperbolas
17.7 Graphing Conic Sections with a Calculator
Chapter 18 Systems of Equations Involving Quadratics
18.1 Graphical Solution
18.2 Algebraic Solution
Chapter 19 Inequalities
19.1 Definitions
19.2 Principles of Inequalities
19.3 Absolute Value Inequalities
19.4 Higher Degree Inequalities
19.5 Linear Inequalities in Two Variables
19.6 Systems of Linear Inequalities
19.7 Linear Programming
Chapter 20 Polynomial Functions
20.1 Polynomial Equations
20.2 Zeros of Polynomial Equations
20.3 Solving Polynomial Equations
20.4 Approximating Real Zeros
Chapter 21 Rational Functions
21.1 Rational Functions
21.2 Vertical Asymptotes
21.3 Horizontal Asymptotes
21.4 Graphing Rational Functions
21.5 Graphing Rational Functions using a Graphing Calculator
Chapter 22 Sequences and Series
22.1 Sequences
22.2 Arithmetic Sequences
22.3 Geometric Sequences
22.4 Infinite Geometric Series
22.5 Harmonic Sequences
22.6 Means
Chapter 23 Logarithms
23.1 Definition of a Logarithm
23.2 Laws of Logarithms
23.3 Common Logarithms
23.4 Using a Common Logarithm Table
23.5 Natural Logarithms
23.6 Using a Natural Logarithm Table
23.7 Finding Logarithms Using a Calculator
Chapter 24 Applications of Logarithms and Exponents
24.1 Introduction
24.2 Simple Interest
24.3 Compound Interest
24.4 Applications of Logarithms
24.5 Applications of Exponents
Chapter 25 Permutations and Combinations
25.1 Fundamental Counting Principle
25.2 Permutations
25.3 Combinations
25.4 Using a Calculator
Chapter 26 The Binomial Theorem
26.1 Combinatorial Notation
26.2 Expansion of (a+x)[sup(n)]
Chapter 27 Probability
27.1 Simple Probability
27.2 Compound Probability
27.3 Mathematical Expectation
27.4 Binomial Probability
27.5 Conditional Probability
Chapter 28 Determinants
28.1 Determinants of Second Order
28.2 Cramer’s Rule
28.3 Determinants of Third Order
28.4 Determinants of Order n
28.5 Properties of Determinants
28.6 Minors
28.7 Value of a Determinant of Order n
28.8 Cramer’s Rule for Determinants of Order n
28.9 Homogeneous Linear Equations
Chapter 29 Matrices
29.1 Definition of a Matrix
29.2 Operations With Matrices
29.3 Elementary Row Operations
29.4 Inverse of a Matrix
29.5 Matrix Equations
29.6 Matrix Solution of a System of Equations
Chapter 30 Mathematical Induction
30.1 Principle of Mathematical Induction
30.2 Proof by Mathematical Induction
Chapter 31 Partial Fractions
31.1 Rational Fractions
31.2 Proper Fractions
31.3 Partial Fractions
31.4 Identically Equal Polynomials
31.5 Fundamental Theorem
31.6 Finding the Partial Fraction Decomposition
Appendix A: Table of Common Logarithms
Appendix B: Table of Natural Logarithms
Index
A
B
C
D
E
F
G
H
I
L
M
N
O
P
Q
R
S
T
U
V
Z

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College Algebra

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College Algebra Fourth Edition Murray R. Spiegel, PhD Former Professor and Chairman, Mathematics Department Rensselaer Polytechnic Institute, Hartford Graduate Center

Robert E. Moyer, PhD Associate Professor of Mathematics Southwest Minnesota State University

Schaum’s Outline Series

New York Chicago San Francisco Athens London Madrid Mexico City Milan New Delhi Singapore Sydney Toronto

MURRAY R. SPIEGEL received the MS degree in Physics and the PhD in Mathematics from Cornell University. He had positions at Harvard University, Columbia University, Oak Ridge, and Rensselaer Polytechnic Institute, and had served as mathematical consultant at several large companies. His last position was Professor and Chairman of Mathematics at the Rensselaer Polytechnic Institute, Hartford Graduate Center. He was interested in most branches of mathematics, especially those which involved applications to physics and engineering problems. He was the author of numerous journal articles and 14 books on various topics in mathematics. DR. ROBERT E. MOYER has been teaching mathematics and mathematics education at Southwest Minnesota State University in Marshall, Minnesota, since 2002. Before coming to SMSU, he taught at Fort Valley State University in Fort Valley, Georgia, from 1985 to 2000, serving as head of the Department of Mathematics and Physics from 1992–1994. Prior to teaching at the university level, Dr. Moyer spent seven years as the mathematics consultant for a five-county Regional Educational Service Agency in central Georgia and twelve years as a high school mathematics teacher in Illinois. He has developed and taught numerous inservice courses for mathematics teachers. He received his Doctor of Philosophy in Mathematics Education from the University of Illinois (Urbana-Champaign) in 1974. He received his Master of Science in 1967 and his Bachelor of Science in 1964, both in Mathematics Education from Southern Illinois University (Carbondale).

Copyright © 2014 by McGraw-Hill Education. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 978-0-07-182585-6 MHID: 0-07-182585-1 The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-182181-0, MHID: 0-07-182181-3. eBook conversion by codeMantra Version 2.0 All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. McGraw-Hill Education eBooks are available at special quantity discounts to use as premiums and sales promotions or for use in corporate training programs. To contact a representative, please visit the Contact Us page at www.mhprofessional.com. Trademarks: McGraw-Hill Education, the McGraw-Hill Education logo, Schaum’s, and related trade dress are trademarks or registered trademarks of McGraw-Hill Education and/or its affiliates in the United States and other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. McGraw-Hill Education is not associated with any product or vendor mentioned in this book. TERMS OF USE This is a copyrighted work and McGraw-Hill Education and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw-Hill Education’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL EDUCATION AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill Education and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill Education nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill Education has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill Education and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.

PREFACE

In the third edition, the comprehensiveness of the second edition is maintained so that all of the topics commonly taught in college algebra are contained in a single source. Recognizing that the use of logarithm tables and determinants is declining, the material on these two areas was reduced, with the two chapters on determinants in the second edition collapsed into a single chapter in the third edition. The material on solving problems using logarithms manually was retained for those who want to learn how to do these problems prior to using a calculator to solve them. Also, the proofs of the properties of determinants were retained to underscore the foundation of the properties used in evaluating determinants. The book is complete in itself and can be used equally well by those who are in a class studying college algebra for the first time as well as those who wish to review the fundamental principles and procedures of college algebra on their own. Students who are studying advanced algebra in high school will be able to use the book as a source of additional examples, explanations, and problems. The thorough treatment of the topics of algebra allows an instructor to use the book as the textbook for a course, as a resource for material on a specific topic, or as a source for additional problems. Each chapter contains a summary of the necessary definitions and theorems followed by a set of solved problems. These solved problems include the proofs of theorems and the derivations of formulas. The chapters end with a set of supplementary problems and their answers. The choice of whether to use a calculator or not is left to the student. A calculator is not required, but it can be used in conjunction with the book. There are no directions on how to use a graphing calculator to do the problems, but there are several instances of the general procedures to be used and the student needs to consult the manual for the calculator being used to see how to implement the procedures on that particular calculator. DR . ROBERT E. MOYER Associate Professor of Mathematics Southwest Minnesota State University

v

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CONTENTS

CHAPTER 1

CHAPTER 2

Fundamental Operations with Numbers

1

1.1 1.2 1.3 1.4 1.5 1.6 1.7

1 2 2 3 3 4 4

Fundamental Operations with Algebraic Expressions 12 2.1 2.2 2.3 2.4 2.5

CHAPTER 3

Special Products Products Yielding Answers of the Form an +bn

22 22 22 23

27 28

32

Factoring Factorization Procedures Greatest Common Factor Least Common Multiple

Fractions 6.1 6.2 6.3

12 12 13 13 13

27

Factoring 5.1 5.2 5.3 5.4

CHAPTER 6

Sets of Numbers Properties Additional Properties

Special Products 4.1 4.2

CHAPTER 5

Algebraic Expressions Terms Degree Grouping Computation with Algebraic Expressions

Properties of Numbers 3.1 3.2 3.3

CHAPTER 4

Four Operations System of Real Numbers Graphical Representation of Real Numbers Properties of Addition and Multiplication of Real Numbers Rules of Signs Exponents and Powers Operations with Fractions

32 33 34 34

41

Rational Algebraic Fractions Operations with Algebraic Fractions Complex Fractions vii

41 42 43

viii

CHAPTER 7

CONTENTS

Exponents 7.1 7.2 7.3 7.4 7.5 7.6

CHAPTER 8

Radicals 8.1 8.2 8.3 8.4 8.5

CHAPTER 9

Equations Operations Used in Transforming Equations Equivalent Equations Formulas Polynomial Equations

Ratio, Proportion, and Variation 11.1 11.2 11.3 11.4 11.5

CHAPTER 12

Complex Numbers Graphical Representation of Complex Numbers Algebraic Operations with Complex Numbers

Equations in General 10.1 10.2 10.3 10.4 10.5

CHAPTER 11

Radical Expressions Laws for Radicals Simplifying Radicals Operations with Radicals Rationalizing Binomial Denominators

Simple Operations with Complex Numbers 9.1 9.2 9.3

CHAPTER 10

Positive Integral Exponent Negative Integral Exponent Roots Rational Exponents General Laws of Exponents Scientific Notation

Ratio Proportion Variation Unit Price Best Buy

Functions and Graphs 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Variables Relations Functions Function Notation Rectangular Coordinate System Function of Two Variables Symmetry

48 48 48 48 49 49 50

58 58 58 58 59 60

67 67 67 68

73 73 73 74 74 75

81 81 81 81 82 82

89 89 89 89 90 90 91 91

CONTENTS

12.8 12.9 12.10

CHAPTER 13

Linear Equations in One Variable 13.1 13.2 13.3

CHAPTER 14

Quadratic Equations Methods of Solving Quadratic Equations Sum and Product of the Roots Nature of the Roots Radical Equations Quadratic-type Equations

Conic Sections 17.1 17.2 17.3 17.4 17.5 17.6 17.7

CHAPTER 18

Systems of Two Linear Equations Systems of Three Linear Equations

Quadratic Equations in One Variable 16.1 16.2 16.3 16.4 16.5 16.6

CHAPTER 17

Slope of a Line Parallel and Perpendicular Lines Slope– Intercept form of Equation of a Line Slope– Point Form of Equation of a Line Two-point Form of Equation of a Line Intercept Form of Equation of a Line

Simultaneous Linear Equations 15.1 15.2

CHAPTER 16

Linear Equations Literal Equations Word Problems

Equations of Lines 14.1 14.2 14.3 14.4 14.5 14.6

CHAPTER 15

Shifts Scaling Using a Graphing Calculator

General Quadratic Equations Conic Sections Circles Parabolas Ellipses Hyperbolas Graphing Conic Sections with a Calculator

Systems of Equations Involving Quadratics 18.1 18.2

Graphical Solution Algebraic Solution

ix

92 93 93

114 114 114 115

128 128 129 130 130 130 131

137 137 138

150 150 150 152 152 152 153

169 169 170 170 171 173 177 180

191 191 191

x

CHAPTER 19

CONTENTS

Inequalities 19.1 19.2 19.3 19.4 19.5 19.6 19.7

CHAPTER 20

Polynomial Functions 20.1 20.2 20.3 20.4

CHAPTER 21

Sequences Arithmetic Sequences Geometric Sequences Infinite Geometric Series Harmonic Sequences Means

Logarithms 23.1 23.2 23.3 23.4 23.5 23.6 23.7

CHAPTER 24

Rational Functions Vertical Asymptotes Horizontal Asymptotes Graphing Rational Functions Graphing Rational Functions using a Graphing Calculator

Sequences and Series 22.1 22.2 22.3 22.4 22.5 22.6

CHAPTER 23

Polynomial Equations Zeros of Polynomial Equations Solving Polynomial Equations Approximating Real Zeros

Rational Functions 21.1 21.2 21.3 21.4 21.5

CHAPTER 22

Definitions Principles of Inequalities Absolute Value Inequalities Higher Degree Inequalities Linear Inequalities in Two Variables Systems of Linear Inequalities Linear Programming

Definition of a Logarithm Laws of Logarithms Common Logarithms Using a Common Logarithm Table Natural Logarithms Using a Natural Logarithm Table Finding Logarithms Using a Calculator

Applications of Logarithms and Exponents 24.1 24.2

Introduction Simple Interest

199 199 199 200 200 202 202 203

214 214 214 216 218

235 235 235 235 236 238

245 245 245 245 246 246 246

263 263 263 264 264 265 265 266

276 276 276

CONTENTS

24.3 24.4 24.5

CHAPTER 25

Permutations and Combinations 25.1 25.2 25.3 25.4

CHAPTER 26

Simple Probability Compound Probability Mathematical Expectation Binomial Probability Conditional Probability

Determinants 28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 28.9

CHAPTER 29

Combinatorial Notation Expansion of (a þ x)n

Probability 27.1 27.2 27.3 27.4 27.5

CHAPTER 28

Fundamental Counting Principle Permutations Combinations Using a Calculator

The Binomial Theorem 26.1 26.2

CHAPTER 27

Compound Interest Applications of Logarithms Applications of Exponents

Determinants of Second Order Cramer’s Rule Determinants of Third Order Determinants of Order n Properties of Determinants Minors Value of a Determinant of Order n Cramer’s Rule for Determinants of Order n Homogeneous Linear Equations

Matrices 29.1 29.2 29.3 29.4 29.5 29.6

Definition of a Matrix Operations With Matrices Elementary Row Operations Inverse of a Matrix Matrix Equations Matrix Solution of a System of Equations

xi

277 278 280

288 288 288 289 290

303 303 303

310 310 310 311 311 311

323 323 323 324 326 327 328 328 328 329

349 349 349 351 352 353 354

xii

CHAPTER 30

CONTENTS

Mathematical Induction 30.1 30.2

CHAPTER 31

Principle of Mathematical Induction Proof by Mathematical Induction

Partial Fractions 31.1 31.2 31.3 31.4 31.5 31.6

Rational Fractions Proper Fractions Partial Fractions Identically Equal Polynomials Fundamental Theorem Finding the Partial Fraction Decomposition

362 362 362

368 368 368 368 369 369 370

APPENDIX A

Table of Common Logarithms

375

APPENDIX B

Table of Natural Logarithms

379

INDEX

383

College Algebra

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CHAPTER 1

Fundamental Operations with Numbers

1.1

FOUR OPERATIONS

Four operations are fundamental in algebra, as in arithmetic. These are addition, subtraction, multiplication, and division. When two numbers a and b are added, their sum is indicated by a þ b. Thus 3 þ 2 ¼ 5. When a number b is subtracted from a number a, the difference is indicated by a  b. Thus 6  2 ¼ 4. Subtraction may be defined in terms of addition. That is, we may define a  b to represent that number x such that x added to b yields a, or x þ b ¼ a. For example, 8  3 is that number x which when added to 3 yields 8, i.e., x þ 3 ¼ 8; thus 8  3 ¼ 5. The product of two numbers a and b is a number c such that a  b ¼ c. The operation of multiplication may be indicated by a cross, a dot or parentheses. Thus 5  3 ¼ 5  3 ¼ 5(3) ¼ (5)(3) ¼ 15, where the factors are 5 and 3 and the product is 15. When letters are used, as in algebra, the notation p  q is usually avoided since  may be confused with a letter representing a number. When a number a is divided by a number b, the quotient obtained is written a 4 b or

a b

or

a=b,

where a is called the dividend and b the divisor. The expression a=b is also called a fraction, having numerator a and denominator b. Division by zero is not defined. See Problems 1.1(b) and (e). Division may be defined in terms of multiplication. That is, we may consider a=b as that number x which upon multiplication by b yields a, or bx ¼ a. For example, 6/3 is that number x such that 3 multiplied by x yields 6, or 3x ¼ 6; thus 6=3 ¼ 2. 1

2

1.2

FUNDAMENTAL OPERATIONS WITH NUMBERS

[CHAP. 1

SYSTEM OF REAL NUMBERS

The system of real numbers as we know it today is a result of gradual progress, as the following indicates. Natural numbers 1, 2, 3, 4, . . . (three dots mean “and so on”) used in counting are also known as the positive integers. If two such numbers are added or multiplied, the result is always a natural number. (2) Positive rational numbers or positive fractions are the quotients of two positive integers, such as 2/3, 8/5, 121/17. The positive rational numbers include the set of natural numbers. Thus the rational number 3/1 is the natural number 3. pffiffiffi (3) Positive irrational numbers are numbers which are not rational, such as 2, p. (4) Zero, written 0, arose in order to enlarge the number system so as to permit such operations as 6  6 or 10  10. Zero has the property that any number multiplied by zero is zero. Zero divided by any number = 0 (i.e., not equal to zero) is zero. pffiffiffi (5) Negative integers, negative rational numbers and negative irrational numbers such as 3, 2=3, and pffiffi2ffi, arose in order to enlarge the number system so as to permit such operations as 2 2 8, p  3p or 2  2 2. pffiffiffi pffiffiffi When no sign is placed before a number, a plus sign is understood. Thus 5 is þ5, 2 is þ 2. Zero is considered a rational number without sign. The real number system consists of the collection of positive and negative rational and irrational numbers and zero. pffiffiffiffiffiffiffi Note. The word “real” is used in contradiction to still other numbers involving 1, which will be taken up later and which are known as imaginary, although they are very useful in mathematics and the sciences. Unless otherwise specified we shall deal with real numbers. (1)

1.3

GRAPHICAL REPRESENTATION OF REAL NUMBERS

It is often useful to represent real numbers by points on a line. To do this, we choose a point on the line to represent the real number zero and call this point the origin. The positive integers þ1, þ2, þ3, . . . are then associated with points on the line at distances 1, 2, 3, . . . units respectively to the right of the origin (see Fig. 1-1), while the negative integers 1, 2, 3, . . . are associated with points on the line at distances 1, 2, 3, . . . units respectively to the left of the origin.

Fig. 1-1

The rational number 1/2 is represented on this scale by a point P halfway between 0 and þ1. The negative number 3=2 or 112 is represented by a point R 112 units to the left of the origin. It can be proved that corresponding to each real number there is one and only one point on the line; and conversely, to every point on the line there corresponds one and only one real number. The position of real numbers on a line establishes an order to the real number system. If a point A lies to the right of another point B on the line we say that the number corresponding to A is greater or larger than the number corresponding to B, or that the number corresponding to B is less or smaller than the number corresponding to A. The symbols for “greater than” and “less than” are . and , respectively. These symbols are called “inequality signs.” Thus since 5 is to the right of 3, 5 is greater than 3 or 5 . 3; we may also say 3 is less than 5 and write 3 , 5. Similarly, since 6 is to the left of 4, 6 is smaller than 4, i.e., 6 , 4; we may also write 4 . 6. By the absolute value or numerical value of a number is meant the distance of the number from the origin on a number line. Absolute value is indicated by two vertical lines surrounding the number. Thus j6j ¼ 6, jþ 4j ¼ 4, j3=4j ¼ 3=4.

4

1.6

FUNDAMENTAL OPERATIONS WITH NUMBERS

[CHAP. 1

EXPONENTS AND POWERS

When a number a is multiplied by itself n times, the product a  a  a    a (n times) is indicated by the symbol an which is referred to as “the nth power of a” or “a to the nth power” or “a to the nth.” EXAMPLES 1.5.

2  2  2  2  2 ¼ 25 ¼ 32, 2  x  x  x ¼ 2x3 ,

(5)3 ¼ (5)(5)(5) ¼ 125

a  a  a  b  b ¼ a3 b2 ,

(a  b)(a  b)(a  b) ¼ (a  b)3

In an , the number a is called the base and the positive integer n is the exponent. If p and q are positive integers, then the following are laws of exponents. (1) ap  aq ¼ apþq ap 1 (2) q ¼ apq ¼ qp a a (3) (ap )q ¼ apq (4) (ab)p ¼ ap bp ,

1.7

if a = 0 ap ap ¼ p b b

if b = 0

Thus: 23  24 ¼ 23þ4 ¼ 27 35 34 1 1 ¼ 352 ¼ 33 ; ¼ ¼ 2 3 36 364 32 (42 )3 ¼ 46 , (34 )2 ¼ 38  3 5 53 2 2 2 (4  5) ¼ 4  5 ¼ 3 2 2

OPERATIONS WITH FRACTIONS

Operations with fractions may be performed according to the following rules. (1)

The value of a fraction remains the same if its numerator and denominator are both multiplied or divided by the same number provided the number is not zero. 3 32 6 ¼ ¼ , 4 42 8

EXAMPLES 1.6.

(2)

Changing the sign of the numerator or denominator of a fraction changes the sign of the fraction. EXAMPLE 1.7.

(3)

3 4 3þ4 7 þ ¼ ¼ 5 5 5 5

The sum or difference of two fractions having different denominators may be found by writing the fractions with a common denominator. EXAMPLE 1.9.

(5)

3 3 3 ¼ ¼ 5 5 5

Adding two fractions with a common denominator yields a fraction whose numerator is the sum of the numerators of the given fractions and whose denominator is the common denominator. EXAMPLE 1.8.

(4)

15 15 4 3 5 ¼ ¼ 18 18 4 3 6

1 2 3 8 11 þ ¼ þ ¼ 4 3 12 12 12

The product of two fractions is a fraction whose numerator is the product of the numerators of the given fractions and whose denominator is the product of the denominators of the fractions. EXAMPLES 1.10.

2 4 24 8  ¼ ¼ , 3 5 3  5 15

3 8 3  8 24 2  ¼ ¼ ¼ 4 9 4  9 36 3

(6)

The reciprocal of a fraction is a fraction whose numerator is the denominator of the given fraction and whose denominator is the numerator of the given fraction. Thus the reciprocal of 3 (i.e., 3/1) is 1/3. Similarly the reciprocals of 5=8 and 4=3 are 8=5 and 3=4 or 3=4, respectively.

(7)

To divide two fractions, multiply the first by the reciprocal of the second. EXAMPLES 1.11.

a c a d ad 4 ¼  ¼ , b d b c bc

2 4 2 5 10 5 4 ¼  ¼ ¼ 3 5 3 4 12 6

CHAP. 1]

FUNDAMENTAL OPERATIONS WITH NUMBERS

5

This result may be established as follows: a c a=b a=b  bd ad 4 ¼ ¼ ¼ : b d c=d c=d  bd bc

Solved Problems 1.1

Write the sum S, difference D, product P, and quotient Q of each of the following pairs of numbers: (a) 48, 12; (b) 8, 0; (c) 0, 12; (d) 10, 20; (e) 0, 0. SOLUTION (a) S ¼ 48 þ 12 ¼ 60, D ¼ 48  12 ¼ 36, P ¼ 48(12) ¼ 576, Q ¼ 48 4 12 ¼

48 ¼4 12

(b) S ¼ 8 þ 0 ¼ 8, D ¼ 8  0 ¼ 8, P ¼ 8(0) ¼ 0, Q ¼ 8 4 0 or 8=0 But by definition 8/0 is that number x (if it exists) such that x(0) ¼ 8. Clearly there is no such number, since any number multiplied by 0 must yield 0. 0 (c) S ¼ 0 þ 12 ¼ 12, D ¼ 0  12 ¼ 12, P ¼ 0(12) ¼ 0, Q ¼ ¼0 12 10 1 (d ) S ¼ 10 þ 20 ¼ 30, D ¼ 10  20 ¼ 10, P ¼ 10(20) ¼ 200, Q ¼ 10 4 20 ¼ ¼ 20 2 (e) S ¼ 0 þ 0 ¼ 0, D ¼ 0  0 ¼ 0, P ¼ 0(0) ¼ 0, Q ¼ 0 4 0 or 0/0 is by definition that number x (if it exists) such that x(0) ¼ 0. Since this is true for all numbers x there is no one number which 0/0 represents. From (b) and (e) it is seen that division by zero is an undefined operation.

1.2

Perform each of the indicated operations. (a) 42 þ 23, 23 þ 42

( f ) 35  28

(i) 72 4 24 þ 64 4 16

(b) 27 þ (48 þ 12), (27 þ 48) þ 12

(g) 756 4 21 (40 þ 21)(72  38) (h) (32  15)

( j) 4 4 2 þ 6 4 3  2 4 2 þ 3  4

(c) 125  (38 þ 27) (d) 6  8, 8  6

(k) 128 4 (2  4), (128 4 2)  4

(e) 4(7  6), (4  7)6 SOLUTION (a) 42 þ 23 ¼ 65, 23 þ 42 ¼ 65. Thus 42 þ 23 ¼ 23 þ 42. This illustrates the commutative law for addition. (b) 27 þ (48 þ 12) ¼ 27 þ 60 ¼ 87, (27 þ 48) þ 12 ¼ 75 þ 12 ¼ 87. Thus 27 þ (48 þ 12) ¼ (27 þ 48) þ 12. This illustrates the associative law for addition. (c) 125  (38 þ 27) ¼ 125  65 ¼ 60 (d ) 6  8 ¼ 48, 8  6 ¼ 48. Thus 6  8 ¼ 8  6, illustrating the commutative law for multiplication. (e) 4(7  6) ¼ 4(42) ¼ 168, (4  7)6 ¼ (28)6 ¼ 168. Thus 4(7  6) ¼ (4  7)6. This illustrates the associative law for multiplication. ( f ) (35)(28) ¼ 35(20 þ 8) ¼ 35(20) þ 35(8) ¼ 700 þ 280 ¼ 980 by the distributive law for multiplication. 756 ¼ 36 Check: 21  36 ¼ 756 21 2 (40 þ 21)(72  38) (61)(34) 61  34 ¼ ¼ ¼ 61  2 ¼ 122 (h) (32  15) 17 17 1

(g)

6

FUNDAMENTAL OPERATIONS WITH NUMBERS

[CHAP. 1

(i) Computations in arithmetic, by convention, obey the following rule: Operations of multiplication and division precede operations of addition and subtraction. Thus 72 4 24 þ 64 4 16 ¼ 3 þ 4 ¼ 7. ( j) The rule of (i) is applied here. Thus 4 4 2 þ 6 4 3  2 4 2 þ 3  4 ¼ 2 þ 2  1 þ 12 ¼ 15. (k) 128 4 (2  4) ¼ 128 4 8 ¼ 16, (128 4 2)  4 ¼ 64  4 ¼ 256 Hence if one wrote 128 4 2 . 4 without parentheses we would do the operations of multiplication and division in the order they occur from left to right, so 128 4 2  4 ¼ 64  4 ¼ 256.

1.3

Classify each of the following numbers according to the categories: real number, positive integer, negative integer, rational number, irrational number, none of the foregoing. pffiffiffi pffiffiffiffiffiffiffi pffiffiffi 5, 3=5, 3p, 2, 1=4, 6:3, 0, 5, 1, 0:3782, 4, 18=7 SOLUTION If the number belongs to one or more categories this is indicated by a check mark.

5 3=5 3p 2 1=4 6.3 0 pffiffiffi 5 pffiffiffiffiffiffiffi 1 0.3782 pffiffiffi 4 18=7

1.4

Real number p

Positive integer

p

Negative integer p

Rational number p

None of foregoing

p

p p

Irrational number

p p

p

p

p

p

p

p

p

p

p p

p p p

p p

p p

Represent (approximately) by a point on a graphical scale each of the real numbers in Problem 1.3. Note: 3p is approximately 3(3.14) ¼ 9.42, so that the corresponding point is between þ9 and þ10 as indicated. pffiffiffi 5 is between 2 and 3, its value to three decimal places being 2.236.

CHAP. 1]

1.5

FUNDAMENTAL OPERATIONS WITH NUMBERS

7

Place an appropriate inequality symbol (, or .) between each pair of real numbers. pffiffiffi (a) 2, 5 (c) 3, 1 (e) 4, 3 (g) p 7,ffiffi3ffi (i) 3=5, 1=2 (b) 0, 2 (d ) 4, þ2 ( f ) p, 3 (h)  2, 1 SOLUTION (a) (b) (c) (d )

1.6

2 , 5 (or 5 . 2), i.e., 2 is less than 5 (or 5 is greater than 2) 0 , 2 (or 2 . 0) (e) 4 , 3 (or 3 . 4) 3 . 1 (or 1 , 3) ( f ) p . 3 (or 3 , p) pffiffiffi pffiffiffi 4 , þ2 (or þ2 . 4) (g) 3 . 7 (or 7 , 3)

Arrange each of the following groups of real numbers in ascending order of magnitude. pffiffiffi pffiffiffi pffiffiffi (a) 3, 22=7, 5, 3:2, 0 (b)  2,  3, 1:6, 3=2 SOLUTION (a) 3:2 , 3 , 0 ,

1.7

pffiffiffi pffiffiffi (h)  2 , 1 (1 .  2) (i) 3=5 , 1=2 since :6 , :5

pffiffiffi 5 , 22=7

pffiffiffi pffiffiffi (b)  3 , 1:6 , 3=2 ,  2

Write the absolute value of each of the following real numbers. pffiffiffi 1, þ3, 2=5,  2, 3:14, 2:83, 3=8, p, þ5=7 SOLUTION We may write the absolute values of these numbers as pffiffiffi j1j, jþ3j, j2=5j, j 2j, j3:14j, j2:83j, j3=8j, jpj, jþ5=7j which in turn may be written 1, 3, 2=5,

1.8

The following illustrate addition and subtraction of real numbers. (a) (3) þ (8) ¼ 11 (b) (2) þ 3 ¼ 1 (c) (6) þ 3 ¼ 3

1.9

pffiffiffi 2, 3:14, 2:83, 3=8, p, 5=7 respectively.

(d) 2 þ 5 ¼ 3 (e) 15 þ 8 ¼ 7 ( f ) (32) þ 48 þ (10) ¼ 6

(g) 50  23  27 ¼ 0 (h) 3  (4) ¼ 3 þ 4 ¼ 1 (i) (14) þ (2) ¼ 14  2 ¼ 12

Write the sum S, difference D, product P, and quotient Q of each of the following pairs of real numbers: (a) 2, 2;

(b) 3, 6;

(c) 0, 5;

(d ) 5, 0

SOLUTION (a) (b) (c) (d )

1.10

S ¼ 2 þ 2 ¼ 0, D ¼ (2)  2 ¼ 4, P ¼ (2)(2) ¼ 4, Q ¼ 2=2 ¼ 1 S ¼ (3) þ 6 ¼ 3, D ¼ (3)  6 ¼ 9, P ¼ (3)(6) ¼ 18, Q ¼ 3=6 ¼ 1=2 S ¼ 0 þ (5) ¼ 5, D ¼ 0  (5) ¼ 5, P ¼ (0)(5) ¼ 0, Q ¼ 0=5 ¼ 0 S ¼ (5) þ 0 ¼ 5, D ¼ (5)  0 ¼ 5, P ¼ (5)(0) ¼ 0, Q ¼ 5=0 (an undefined operation, so it is not a number).

Perform the indicated operations. (a) (5)(3)(2) ¼ ½(5)(3)(2) ¼ (15)(2) ¼ 30 ¼ (5)½(3)(2) ¼ (5)(6) ¼ 30 The arrangement of the factors of a product does not affect the result.

8

FUNDAMENTAL OPERATIONS WITH NUMBERS

[CHAP. 1

(b) 8(3)(10) ¼ 240

1.11

(c)

8(2) (4)(2) 16 8 þ ¼ þ ¼4þ4¼8 4 2 4 2

(d)

12(40)(12) 12(40)(12) 12(40)(12) ¼ ¼ ¼ 960 5(3)  3(3) 15  (9) 6

Evaluate the following. (a) 23 ¼ 2  2  2 ¼ 8 (b) 5(3)2 ¼ 5  3  3 ¼ 45 (c) 24  26 ¼ 24þ6 ¼ 210 ¼ 1024 (d ) 25  52 ¼ (32)(25) ¼ 800 (e)

34  33 37 ¼ 2 ¼ 372 ¼ 35 ¼ 243 32 3

(f)

52  53 55 1 1 1 ¼ 7 ¼ 75 ¼ 2 ¼ 5 5 25 57 5

(g) (23 )2 ¼ 232 ¼ 26 ¼ 64

1.12

(h)

 4 2 24 16 ¼ 4¼ 3 81 3

(i)

(34 )3  (32 )4 312  38 320 ¼ ¼  19 ¼ 31 ¼ 3 15 15 4 4 3  3 3 (3)  3

( j)

38 42  24 42  þ 3(2)3 ¼ 33  2 þ 3(8) ¼ 27  4  24 ¼ 1 5 6 3 2 2

Write each of the following fractions as an equivalent fraction having the indicated denominator. (a) 1/3; 6

(b) 3/4; 20

(c) 5/8; 48

(d ) 3=7; 63

(e) 12=5; 75

SOLUTION (a) To obtain the 1 1 ¼ Then 3 3

1.13

denominator 6, multiply numerator and denominator of the fraction 1/3 by 2. 2 2  ¼ . 2 6

(b)

3 3  5 15 ¼ ¼ 4 4  5 20

3 39 27 (d )  ¼  ¼ 7 79 63

(c)

5 5  6 30 ¼ ¼ 8 8  6 48

(e) 

12 12  15 180 ¼ ¼ 5 5  15 75

Find the sum S, difference D, product P, and quotient Q of each of the following pairs of rational numbers: (a) 1=3, 1=6; (b) 2=5, 3=4; (c) 4=15, 11=24.

CHAP. 1]

FUNDAMENTAL OPERATIONS WITH NUMBERS

SOLUTION (a) 1=3 may be written as the equivalent fraction 2=6. S¼

1 1 2 1 3 1 þ ¼ þ ¼ ¼ 3 6 6 6 6 2

P¼

   1 1 1 ¼ 3 6 18

D¼

1 1 2 1 1  ¼  ¼ 3 6 6 6 6

Q¼

1=3 1 6 6 ¼  ¼ ¼2 1=6 3 1 3

(b) 2=5 and 3=4 may be expressed with denominator 20: 2=5 ¼ 8=20, 3=4 ¼ 15=20. S¼

2 3 8 15 23 þ ¼ þ ¼ 5 4 20 20 20

P¼

   2 3 6 3 ¼ ¼ 5 4 20 10

D¼

2 3 8 15 7  ¼  ¼ 5 4 20 20 20

Q¼

2=5 2 4 8 ¼  ¼ 3=4 5 3 15

(c)  4=15 and 11=24 have a least common denominator 120: 4=15 ¼ 32=120, 11=24 ¼ 55=120.     4 11 32 55 87 29  ¼ ¼ S¼  þ  ¼ 15 24 120 120 120 40     4 11 32 55 23   ¼ þ ¼ D¼  15 24 120 120 120

1.14

   4 11 11 P¼   ¼ 15 24 90    4=15 4 24 32 Q¼ ¼   ¼ 11=24 15 11 55

Evaluate the following expressions, given x ¼ 2, y ¼ 3, z ¼ 5, a ¼ 1=2, b ¼ 2=3. (a) 2x þ y ¼ 2(2) þ (3) ¼ 4  3 ¼ 1 (b) 3x  2y  4z ¼ 3(2)  2(3)  4(5) ¼ 6 þ 6  20 ¼ 8 (c) 4x2 y ¼ 4(2)2 (3) ¼ 4  4  (3) ¼ 48 (d)

x3 þ 4y 23 þ 4(3) 8  12 4 ¼ ¼ ¼ 2a  3b 2(1=2)  3(2=3) 1þ2 3

(e)

 2  3  2    2  3   x b 2 2=3 3 2 4 4 64 4 64 68 ¼ þ ¼ 3 ¼ 3  3  ¼ 3  y a 3 1=2 3 3 9 27 9 9 9

Supplementary Problems 1.15

Write the sum S, difference D, product P, and quotient Q of each of the following pairs of numbers: (a) 54, 18; (b) 4, 0; (c) 0, 4; (d) 12, 24; (e) 50, 75.

1.16

Perform each of the indicated operations. (a) (b) (c) (d ) (e) (f) (g)

38 þ 57, 57 þ 38 15 þ (33 þ 8), (15 þ 33) þ 8 (23 þ 64)  (41 þ 12) 12  8, 8  12 6(4  8), (6  4)8 42  68 1296 4 36

(h)

(35  23)(28 þ 17) 43  25

(i) 45 4 15 þ 84 4 12 ( j) 10 4 5  4 4 2 þ 15 4 3 þ 2  5 (k) 112 4 (4  7), (112 4 4)  7 15 þ 3  2 (l ) 9442

9

10

FUNDAMENTAL OPERATIONS WITH NUMBERS

1.17

Place an appropriate inequality symbol (, or .) between each of the following pairs of real numbers. pffiffiffiffiffi (a) 4, 3 (c) 1, 2 (e) 8, (g) 3,  11 pffiffi7 ffi (b) 2, 0 (d ) 3, 2 ( f ) 1, 2 (h) 1=3, 2=5

1.18

Arrange each of the following groups of real numbers in ascending order of magnitude. pffiffiffi pffiffiffi pffiffiffi (a)  3, 2, 6, 2:8, 4, 7=2 (b) 2p, 6, 8, 3p, 4:8, 19=3

[CHAP. 1

1.19

pffiffiffi pffiffiffi Write the absolute value of each of the following real numbers: 2, 3=2,  6, þ3:14, 0, 5=3, 4, 0:001, p  1.

1.20

Evaluate.

1.21

(a) 6 þ 5

(d ) 6 þ (4)

(g) (18) þ (3) þ 22

(b) (4) þ (6)

(e) 8 þ 4

(h) 40  12 þ 4

(c) (4) þ 3

( f ) 4 þ 8

(i) 12  (8)

( j) (16)  (12) þ (5)  15

Write the sum S, difference D, product P, and quotient Q of each of the following pairs of real numbers: (a) 12, 4; (b) 6, 3; (c) 8, 4; (d ) 0, 4; (e) 3, 2.

1.22

1.23

Perform the indicated operations. (a) (3)(2)(6)

(c) 4(1)(5) þ (3)(2)(4)

(b) (6)(8)(2)

(d )

(4)(6) (16)(9) þ 3 12

Evaluate.

(e) (8) 4 (4) þ (3)(2) (f)

(3)(8)(2) (4)(6)  (2)(12)

(a) 33

(e)

56  53 55

(i)

 6 1  25 2

(b) 3(4)2

(f )

34  38 36  35

( j)

(2)3  (2)3 3(22 )2

(c) 24  23

(g)

75 73  74

(k)

3(3)2 þ 4(2)3 23  32

(d ) 42  32

(h) (32 )3

(l)

57 210 þ 2  4(3)4 4 5 8  (2)3

1.24

Write each of the following fractions as an equivalent fraction having the indicated denominator. (a) 2/5; 15 (c) 5/16; 64 (e) 11/12; 132 (b) 24/7; 28 (d ) 10=3; 42 ( f ) 17/18; 90

1.25

Find the sum S, difference D, product P, and quotient Q of each of the following pairs of rational numbers: (a) 1/4, 3/8; (b) 1/3, 2/5; (c) 24, 2/3; (d ) 22/3, 3=2.

1.26

Evaluate the following expressions, given x ¼ 2, y ¼ 4, z ¼ 1=3, a ¼ 1, b ¼ 1=2: (a) 3x  2y þ 6z

(d )

(b) 2xy þ 6az

(e)

(c) 4b2 x3

3y2  4x ax þ by

x2 y(x þ y) 3x þ 4y  3  2 y a xy (f) 4  2 x b z

CHAP. 1]

FUNDAMENTAL OPERATIONS WITH NUMBERS

ANSWERS TO SUPPLEMENTARY PROBLEMS 1.15

(a) S ¼ 72, D ¼ 36, P ¼ 972, Q ¼ 3 (b) S ¼ 4, D ¼ 4, P ¼ 0, Q undefined (c) S ¼ 4, D ¼ 4, P ¼ 0, Q ¼ 0

(d) S ¼ 36, D ¼ 12, P ¼ 288, Q ¼ 1=2 (e) S ¼ 125, D ¼ 25, P ¼ 3750, Q ¼ 2=3

1.16

(a) 95, 95 (b) 56, 56

(g) 36 (h) 30

1.17

(a) 3 , 4 or 4 . 3 (b) 2 , 0 or 0 . 2 (c) 1 , 2 or 2 . 1

(c) 34 (d ) 96, 96

(e) 192, 192 ( f ) 2856

(i) 10 ( j) 15

(d ) 2 , 3 or 3 . 2 (e) 8 , p7 ffiffiffi orp7 ffiffiffi . 8 ( f ) 1 , 2 or 2 . 1

(k) 4, 196 (l) 3

pffiffiffiffiffi pffiffiffiffiffi (g)  11 , 3 or 3 .  11 (h) 2=5 , 1=3 or 1=3 . 2=5

1.18

pffiffiffi pffiffiffi (a) 2:8 , 2 ,  3 , 6 , 7=2 , 4

1.19

2, 3=2,

1.20

(a) 11 (b) 10

1.21

(a) S ¼ 16, D ¼ 8, P ¼ 48, Q ¼ 3 (b) S ¼ 9, D ¼ 3, P ¼ 18, Q ¼ 2 (c) S ¼ 4, D ¼ 12, P ¼ 32, Q ¼ 2

1.22

(a) 36

(b) 96

1.23

(a) 27 (b) 48

(c) 128 (d ) 144

1.24

(a) 6=15 (b) 16=28 (c) 20=64 (d ) 140=42 (e) 121=132

1.25

(a) (b) (c) (d )

1.26

(a) 12 (b) 18 (c) 8 (d ) 14 (e) 16=5

(b) 3p , 6 ,

pffiffiffi 8 , 4:8 , 2p , 19=3

pffiffiffi pffiffiffi 6, 3:14, 0, 5=3, 4, 0:001, p þ 1 (c) 1 (d ) 2

(e) 4 (f ) 4

(c) 4

(d) 20

(e) 54 ¼ 625 (f) 3

(i) 4 ( j) 8

(g) 1 (h) 32

(d ) S ¼ 4, D ¼ 4, P ¼ 0, Q ¼ 0 (e) S ¼ 1, D ¼ 5, P ¼ 6, Q ¼ 3=2

(e) 4

(f) 1

(g) 1=49 (h) 36 ¼ 729

(i) 1=2 ( j) 4=3

S ¼ 5=8, D ¼ 1=8, P ¼ 3=32, Q ¼ 2=3 S ¼ 11=15, D ¼ 1=15, P ¼ 2=15, Q ¼ 5=6 S ¼ 10=3, D ¼ 14=3, P ¼ 8=3, Q ¼ 6 S ¼ 13=6, D ¼ 5=6, P ¼ 1, Q ¼ 4=9 ( f ) 48

(k) 5 (l) 201 ( f ) 85=90

11

CHAPTER 2

Fundamental Operations with Algebraic Expressions 2.1

ALGEBRAIC EXPRESSIONS

An algebraic expression is a combination of ordinary numbers and letters which represent numbers. Thus

3x2  5xy þ 2y4 ,

2a3 b5 ,

5xy þ 3z 2a3  c2

are algebraic expressions. A term consists of products and quotients of ordinary numbers and letters which represent numbers. Thus 6x2 y3 , 5x=3y4 , 3x7 are terms. However, 6x2 þ 7xy is an algebraic expression consisting of two terms. A monomial is an algebraic expression consisting of only one term. Thus 7x3 y4 , 3xyz2 , 4x2=y are monomials. Because of this definition, monomials are sometimes simply called terms. A binomial is an algebraic expression consisting of two terms. Thus 2x þ 4y, 3x4  4xyz3 are binomials. A trinomial is an algebraic expression consisting of three terms. Thus 3x2  5x þ 2, 2x þ 6y  3z, 3 x  3xy=z  2x3 z7 are trinomials. A multinomial is an algebraic expression consisting of more than one term. Thus 7x þ 6y, 3x3 þ 6x2 y  7xy þ 6, 7x þ 5x2 =y  3x3 =16 are multinomials. 2.2

TERMS

One factor of a term is said to be the coefficient of the rest of the term. Thus in the term 5x3 y2 , 5x3 is the coefficient of y2 , 5y2 is the coefficient of x3 , and 5 is the coefficient of x3 y2 . If a term consists of the product of an ordinary number and one or more letters, we call the number the numerical coefficient (or simply the coefficient) of the term. Thus in 5x3 y2 , 5 is the numerical coefficient or simply the coefficient. 12

CHAP. 2]

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

13

Like terms, or similar terms, are terms which differ only in numerical coefficients, For example, 7xy and 2xy are like terms; 3x2 y4 and  12 x2 y4 are like terms; however, 2a2 b3 and 3a2 b7 are unlike terms. Two or more like terms in an algebraic expression may be combined into one term. Thus 7x2 y  4x2 y þ 2 2x y may be combined and written 5x2 y. A term is integral and rational in certain literals (letters which represent numbers) if the term consists of (a) positive integer powers of the variables multiplied by a factor not containing any variable, or (b) no variables at all.

pffiffiffi 3 6 2 3 4 For example, the terms 6x y , 5y , 7, 4x, and 3x y are integral and rational in the variables present. pffiffiffi However, 3 x is not rational in x, 4=x is not integral in x. A polynomial is a monomial or multinomial in which every term is integral and rational. For example, 3x2 y3  5x4 y þ 2, 2x4  7x3 þ 3x2  5x þ 2, 4xy þ z, and 3x2 are polynomials. However, pffiffiffi 3x2  4=x and 4 y þ 3 are not polynomials.

2.3

DEGREE

3 2 The degree of a monomial is the sum of all the exponents in the pffiffiffi variables in the term. Thus the degree of 4x y z is 3 þ 2 þ 1 ¼ 6. The degree of a constant, such as 6, 0,  3, or p, is zero. The degree of a polynomial is the same as that of the term having highest degree and non-zero coefficient. Thus 7x3 y2  4xz5 þ 2x3 y has terms of degree 5, 6, and 4 respectively; hence the degree of the polynomial is 6.

2.4

GROUPING

A symbol of grouping such as parentheses ( ), brackets [ ], or braces { } is often used to show that the terms contained in them are considered as a single quantity. For example, the sum of two algebraic expressions 5x2  3x þ y and 2x  3y may be written 2 (5x  3x þ y) þ (2x  3y). The difference of these may be written (5x2  3x þ y)  (2x  3y), and their product (5x2  3x þ y)(2x  3y). Removal of symbols of grouping is governed by the following laws. (1)

If a þ sign precedes a symbol of grouping, this symbol of grouping may be removed without affecting the terms contained. Thus

(2)

If a 2 sign precedes a symbol of grouping, this symbol of grouping may be removed if each sign of the terms contained is changed. Thus

(3)

(3x þ 7y)  (4xy  3x3 ) ¼ 3x þ 7y  4xy þ 3x3 :

If more than one symbol of grouping is present, the inner ones are to be removed first. Thus

2.5

(3x þ 7y) þ (4xy  3x3 ) ¼ 3x þ 7y þ 4xy  3x3 :

2x  {4x3  (3x2  5y)} ¼ 2x  {4x3  3x2 þ 5y} ¼ 2x  4x3 þ 3x2  5y:

COMPUTATION WITH ALGEBRAIC EXPRESSIONS

Addition of algebraic expressions is achieved by combining like terms. In order to accomplish this addition, the expressions may be arranged in rows with like terms in the same column; these columns are then added.

14

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

EXAMPLE 2.1.

[CHAP. 2

Add 7x þ 3y3  4xy, 3x  2y3 þ 7xy, and 2xy  5x  6y3 . 7x 3y 3 24xy 3x 22y 3 7xy 2xy 25x 26y 3 ___________________ 3 Addition: 5x 25y 5xy. Write:

Hence the result is 5x 2 5y 3 þ 5xy.

Subtraction of two algebraic expressions is achieved by changing the sign of every term in the expression which is being subtracted (sometimes called the subtrahend) and adding this result to the other expression (called the minuend). EXAMPLE 2.2.

Subtract 2x2  3xy þ 5y2 from 10x2  2xy  3y2 .

Subtraction:

10x 2 2 2xy 2 3y 2 2x 2 2 3xy þ 5y 2 _______________ 8x 2 þ xy 2 8y 2

We may also write (10x 2 2 2xy 2 3y 2) 2 (2x 2 2 3xy þ 5y 2) ¼ 10x 2 2 2xy 2 3y 2 2 2x 2 þ 3xy 2 5y 2 ¼ 8x 2 þ xy 2 8y 2. Multiplication of algebraic expressions is achieved by multiplying the terms in the factors of the expressions. (1)

To multiply two or more monomials: Use the laws of exponents, the rules of signs, and the commutative and associative properties of multiplication. EXAMPLE 2.3. Multiply 3x2 y3 z, 2x4 y, and 4xy4 z2 . Write

(3x2 y3 z)(2x4 y)(4xy4 z2 ):

Arranging according to the commutative and associative laws, {(3)(2)(4)g{(x2 )(x4 )(x)}{(y3 )(y)(y4 )}{(z)(z2 )}:

(1)

Combine using rules of signs and laws of exponents to obtain 24x7 y8 z3 : Step (1) may be done mentally when experience is acquired.

(2)

To multiply a polynomial by a monomial: Multiply each term of the polynomial by the monomial and combine results. EXAMPLE 2.4. Multiply 3xy  4x3 þ 2xy2 by 5x2 y4 . Write

(5x2 y4 )(3xy  4x3 þ 2xy2 ) ¼ (5x2 y4 )(3xy) þ (5x2 y4 )(4x3 ) þ (5x2 y4 )(2xy2 ) ¼ 15x3 y5  20x5 y4 þ 10x3 y6 :

(3)

To multiply a polynomial by a polynomial: Multiply each of the terms of one polynomial by each of the terms of the other polynomial and combine results. It is very often useful to arrange the polynomials according to ascending (or descending) powers of one of the letters involved.

CHAP. 2]

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

EXAMPLE 2.5. Multiply 3x þ 9 þ x2 by 3  x. Arranging in descending powers of x, x 2 2 3x þ 9 2x þ 3 __________ Multiplying (2) by 2x, 2x 3 þ 3x 2 2 9x Multiplying (2) by 3, 3x 2 2 9x þ 27 ___________________ Adding, 2x 3 þ 6x 2 2 18x þ 27

15

(2)

Division of algebraic expressions is achieved by using the division laws of exponents. (1)

To divide a monomial by a monomial: Find the quotient of the numerical coefficients, find the quotients of the variables, and multiply these quotients. EXAMPLE 2.6. Write

(2)

Divide 24x4 y2 z3 by 3x3 y4 z.   4  2  3    24x4 y2 z3 24 x y z 1 8xz2 ¼ ¼ (8)(x) (z2 ) ¼  2 : 3 4 3 4 2 3 x y 3x y z y z y

To divide a polynomial by a polynomial: (a) Arrange the terms of both polynomials in descending (or ascending) powers of one of the variables common to both polynomials. (b) Divide the first term in the dividend by the first term in the divisor. This gives the first term of the quotient. (c) Multiply the first term of the quotient by the divisor and subtract from the dividend, thus obtaining a new dividend. (d ) Use the dividend obtained in (c) to repeat steps (b) and (c) until a remainder is obtained which is either of degree lower than the degree of the divisor or zero. (e) The result is written: dividend remainder ¼ quotient þ : divisor divisor

EXAMPLE 2.7.

Divide x2 þ 2x4  3x3 þ x  2 by x2  3x þ 2.

Write the polynomials in descending powers of x and arrange the work as follows. 2x 2 þ 3x þ 6 ) x 2 3x þ 2 2x4  3x3 þ x2 þ x  2 2x 4 2 6x 3 þ 4x 2 3x3  3x2 þ x  2 3x3  9x2 þ 6x 6x2  5x  2 6x2  18x þ 12 13x  14 2

Hence

2x4  3x3 þ x2 þ x  2 13x  14 ¼ 2x2 þ 3x þ 6 þ 2 : x2  3x þ 2 x  3x þ 2

16

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

[CHAP. 2

Solved Problems 2.1

Evaluate each of the following algebraic expressions, given that x ¼ 2, y ¼ 1, z ¼ 3, a ¼ 0, b ¼ 4, c ¼ 1=3: (a) 2x2  3yz ¼ 2(2)2  3(1)(3) ¼ 8 þ 9 ¼ 17 (b) 2z4  3z3 þ 4z2  2z þ 3 ¼ 2(3)4  3(3)3 þ 4(3)2  2(3) þ 3 ¼ 162  81 þ 36  6 þ 3 ¼ 114 (c) 4a2  3ab þ 6c ¼ 4(0)2  3(0)(4) þ 6(1=3) ¼ 0  0 þ 2 ¼ 2

2.2

(d )

5xy þ 3z 5(2)(1) þ 3(3) 10 þ 9 1 ¼ ¼9 ¼ ¼ 3 2 3 2 2a  c 1=9 1=9 2(0)  (1=3)

(e)

3x2 y bc 3(2)2 (1) 4(1=3)  ¼  ¼ 4  4=9 ¼ 40=9 z xþ1 3 3

(f )

4x2 y(z  1) 4(2)2 (1)(3  1) 4(4)(1)(2) 32 ¼ ¼ ¼ a þ b  3c 0 þ 4  3(1=3) 41 3

Classify each of the following algebraic expressions according to the categories: term or monomial, binomial, trinomial, multinomial, polynomial. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d ) y þ 3 (g) x2 þ y2 þ z2 (a) x3 þ 3y2 z pffiffi pffiffiffi pffiffi (b) 2x2  5x þ 3 (e) 4z2 þ 3z  2 z (h) y þ z 2 3 ( f ) 5x þ 4=y (i) a3 þ b3 þ c3  3abc (c) 4x y=z SOLUTION If the expression belongs to one or more categories, this is indicated by a check mark. Term or monomial x þ 3y z 3

2

2x  5x þ 3

Binomial p

Trinomial

Multinomial p

Polynomial p

p

p

p

p

p

2

4x2 y=z

p p

yþ3 pffiffi 2 4z þ 3z  2 z 5x3 þ 4=y pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z2 pffiffiffi pffiffi yþ z

p p

p

p

p

p

a3 þ b3 þ c3  3abc

2.3

p

p

p

Find the degree of each of the following polynomials. (a) 2x3 y þ 4xyz4 . The degree of 2x3 y is 4 and that of 4xyz4 is 6; hence the polynomial is of degree 6. (b) x2 þ 3x3  4. The degree of x2 is 2, of 3x3 is 3, and of 24 is 0; hence the degree of the polynomial is 3. (c) y3  3y2 þ 4y  2 is of degree 3. (d ) xz3 þ 3x2 z2  4x3 z þ x4 . Each term is of degree 4; hence the polynomial is of degree 4. (e) x2  105 is of degree 2. (The degree of the constant 105 is zero.)

CHAP. 2]

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

17

2.4

Remove the symbols of grouping in each of the following and simplify the resulting expressions by combining like terms. (a) 3x2 þ (y2  4z)  (2x  3y þ 4z) ¼ 3x2 þ y2  4z  2x þ 3y  4z ¼ 3x2 þ y2  2x þ 3y  8z (b) 2(4xy þ 3z) þ 3(x  2xy)  4(z  2xy) ¼ 8xy þ 6z þ 3x  6xy  4z þ 8xy ¼ 10xy þ 3x þ 2z (c) x  3  2{2  3(x  y)} ¼ x  3  2{2  3x þ 3y} ¼ x  3  4 þ 6x  6y ¼ 7x  6y  7 (d) 4x2  {3x2  2½y  3(x2  y) þ 4} ¼ 4x2  {3x2  2½y  3x2 þ 3y þ 4} ¼ 4x2  {3x2  2y þ 6x2  6y þ 4} ¼ 4x2  {9x2  8y þ 4} ¼ 4x2  9x2 þ 8y  4 ¼ 5x2 þ 8y  4

2.5

Add the algebraic expressions in each of the following groups. (a) x2 þ y2  z2 þ 2xy  2yz, 1  x2  y2  z2

y2 þ z2  x2 þ 2yz  2zx,

z2 þ x2  y2 þ 2zx  2xy,

SOLUTION Arranging,

Adding,

x2 þ y2  z2 þ 2xy  2yz x2 þ y2 þ z2 þ 2yz  2zx x2  y2 þ z2  2xy þ 2zx þ1 x2  y2  z2 0 þ0 þ0 þ0

(b) 5x3 y  4ab þ c2 ,

þ0

þ0

3c2 þ 2ab  3x2 y,

þ1

The result of the addition is 1.

x3 y þ x2 y  4c2  3ab,

4c2  2x2 y þ ab2  3ab

SOLUTION

2.6

Arranging,

5x3 y  4ab þ c2 3x2 y þ 2ab þ 3c2 x2 y þ x3 y  3ab  4c2 2x2 y  3ab þ 4c2 þ ab2

Adding,

4x2 y þ 6x3 y  8ab þ 4c2 þ ab2

Subtract the second of each of the following expressions from the first. (a) a  b þ c  d, c  a þ d  b: SOLUTION Write

abþc d a  b þ c þ d

Subtracting,

2a þ 0 þ 0  2d

The result is 2a  2d:

Otherwise: (a  b þ c  d)  (c  a þ d  b) ¼ a  b þ c  d  c þ a  d þ b ¼ 2a  2d

(b) 4x2 y  3ab þ 2a2  xy,

4xy þ ab2  3a2 þ 2ab:

SOLUTION Write

4x2 y  3ab þ 2a2  xy 2ab  3a2 þ 4xy þ ab2

Subtracting,

4x2 y  5ab þ 5a2  5xy  ab2

Otherwise: (4x2 y  3ab þ 2a2  xy)  (4xy þ ab2  3a2 þ 2ab)

¼ 4x2 y  3ab þ 2a2  xy  4xy  ab2 þ 3a2  2ab ¼ 4x2 y  5ab þ 5a2  5xy  ab2

18

2.7

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

In each of the following find the indicated product of the algebraic expressions. (a) (b) (c) (d )

(2ab3 )(4a2 b5 ) (3x2 y)(4xy2 )(2x3 y4 ) (3ab2 )(2ab þ b2 ) (x2  3xy þ y2 )(4xy2 )

(e) (f ) (g) (h)

(x2  3x þ 9)(x þ 3) (x4 þ x3 y þ x2 y2 þ xy3 þ y4 )(x  y) (x2  xy þ y2 )(x2 þ xy þ y2 ) (2x þ y  z)(3x  z þ y)

SOLUTION (a) (2ab3 )(4a2 b5 ) ¼ {(2)(4)}{(a)(a2 )}{(b3 )(b5 )} ¼ 8a3 b8 (b) (3x2 y)(4xy2 )(2x3 y4 ) ¼ {(3)(4)(2)}{(x2 )(x)(x3 )}{(y)(y2 )(y4 )} ¼ 24x6 y7 (c) (3ab2 )(2ab þ b2 ) ¼ (3ab2 )(2ab) þ (3ab2 )(b2 ) ¼ 6a2 b3 þ 3ab4 (d ) (x2  3xy þ y2 )(4xy2 ) ¼ (x2 )(4xy2 ) þ (3xy)(4xy2 ) þ (y2 )(4xy2 ) ¼ 4x3 y2  12x2 y3 þ 4xy4 (e) x2  3x þ 9 xþ3 x3  3x2 þ 9x 3x2  9x þ 27

x5 þ x4 y þ x3 y2 þ x2 y3 þ xy4 x4 y  x3 y2  x2 y3  xy4  y5

x3 þ 0 þ 0 þ 27 Ans: x3 þ 27

x5 þ 0 þ 0 þ 0 þ 0  y5 Ans: x5  y5

(g) x2  xy þ y2 x2 þ xy þ y2 x4  x3 y þ x2 y2 x3 y  x2 y2 þ xy3 x2 y2  xy3 þ y4 x4 þ 0 þ x2 y2 þ 0 þ y4 Ans: x4 þ x2 y2 þ y4

2.8

( f ) x4 þ x3 y þ x2 y2 þ xy3 þ y4 xy

(h) 2x þ y  z 3x þ y  z 6x2 þ 3xy  3xz 2xy þ y2  yz  2xz  yz þ z2 6x2 þ 5xy  5xz þ y2 2yz þ z2

Perform the indicated divisions.   3  2     24x3 y2 z 24 x y z 1 6x2 y 2 ¼ ¼ (6)(x ¼ )(y) 2 2 4xyz 4 z z z x y         16a4 b6 16 a4 b6 1 2a3 b4 ¼ (b) ¼ 8 8ab2 c a b2 c c       3x3 y þ 16xy2  12x4 yz4 3x3 y 16xy2 12x4 yz4 3x 8y (c) ¼ þ þ ¼ þ  6x2 z3 2 2 2 2 2x yz 2z xz 2x yz 2x yz 2x yz  3 2     3 2 2 2 4a b þ 16ab  4a 4a b 16ab 4a 8 2 ¼ (d ) þ þ ¼ 2ab  þ 2a2 b a b 2a2 b 2a2 b 2a2 b

(a)

(e)

2x4 þ 3x3  x2  1 x2 2x3 þ 7x2 þ 13x þ 26 ) 1 x  2 2x4 þ 3x3  x2 2x4  4x3 7x3  x2 1 7x3  14x2 13x2 1 13x2  26x 26x  1 26x  52 51

(f)

16y4  1 2y  1 8y3 þ 4y2 þ 2y þ 1 ) 2y  1 16y4 1 16y4  8y3 8y3 1 8y3  4y2 4y2 1 4y2  2y 2y  1 2y  1 0

[CHAP. 2

CHAP. 2]

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

Thus (g)

2x4 þ 3x3  x2  1 51 16y4  1 ¼ 2x3 þ 7x2 þ 13x þ 26 þ and ¼ 8y3 þ 4y2 þ 2y þ 1. x2 x2 2y  1

2x6 þ 5x4  x3 þ 1 . x2 þ x þ 1

Arrange in descending powers of x. 2x4  2x3  9x2  10x  19 þ 5x4  x3 þ1 x þ x þ 1)2x6 6 5 2x  2x  2x4 2x5 þ 7x4  x3 þ1 2x5  2x4  2x3 9x4 þ x3 þ1 9x4  9x3  9x2 10x3 þ 9x2 þ1 10x3  10x2  10x 19x2 þ 10x þ 1 19x2  19x  19 29x þ 20 2

Thus

(h)

2x6 þ 5x4  x3 þ 1 29x þ 20 ¼ 2x4  2x3  9x2  10x  19 þ 2 . x2 þ x þ 1 x þ x þ 1

x4  x3 y þ x2 y2 þ 2x2 y  2xy2 þ 2y3 . x2  xy þ y2

Arrange in descending powers of a letter, say x. x  xy þ 2

Thus

2.9

x2 þ 2y  x3 y þ x2 y2 þ 2x2 y  2xy2 þ 2y3 4 x  x3 y þ x2 y2 2x2 y  2xy2 þ 2y3 2x2 y  2xy2 þ 2y3 0

y2 ) x4

x4  x3 y þ x2 y2 þ 2x2 y  2xy2 þ 2y3 ¼ x2  2y. x2  xy þ y2

Check the work in Problems 2.7(h) and 2.8(g) by using the value x ¼ 1, y ¼ 1, z ¼ 2. SOLUTION From Problem 2.7(h), (2x þ y  z)(3x  z þ y) ¼ 6x2 þ 5xy  5xz  2yz þ z2 þ y2 : Substitute x ¼ 1, y ¼ 1, z ¼ 2 and obtain ½2(1) þ (1)  2½3(1)  (2)  1 ¼ 6(1)2 þ 5(1)(1)  5(1)(2)  2(1)(2) þ (2)2 þ (1)2 ½1½0 ¼ 6  5  10 þ 4 þ 4 þ 1, i.e. 0 ¼ 0:

or From Problem 2.8(g),

2x6 þ 5x4  x3 þ 1 29x þ 20 ¼ 2x4  2x3  9x2  10x  19 þ 2 : x2 þ x þ 1 x þ x þ 1

19

20

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

[CHAP. 2

Put x ¼ 1 and obtain 2þ51þ1 29 þ 20 ¼ 2  2  9  10  19 þ 1 þ 1 þ 1 1 þ 1 þ 1

or

7 ¼ 7:

Although a check by substitution of numbers for variables is not conclusive, it can be used to indicate possible errors.

Supplementary Problems 2.10

Evaluate each algebraic expression, given that x ¼ 1, y ¼ 3, z ¼ 2, a ¼ 1=2, b ¼ 2=3. (a) 4x3 y2  3xz2

(e)

z(x þ y) 3ab  2 8a yxþ1

(g)

1 1 1 þ þ x y z

(f )

(x  y)2 þ 2z ax þ by

(h)

(x  1)(y  1)(z  1) (a  1)(b  1)

(b) (x  y)(y  z)(z  x) (c) 9ab2 þ 6ab  4a2 (d ) 2.11

2.12

2.13

xy2  3z aþb

Determine the degree of each of the following polynomials. (a) 3x4  2x3 þ x2  5

(c) x5 þ y5 þ z5  5xyz

(b) 4xy4  3x3 y3

(d )

(e) 103

pffiffiffi 3xyz  5

( f ) y2  3y5  y þ 2y3  4

Remove the symbols of grouping and simplify the resulting expressions by combining like terms. (a) (x þ 3y  z)  (2y  x þ 3z) þ (4z  3x þ 2y)

(c) 3x þ 4y þ 3{x  2(y  x)  y}

(b) 3(x2  2yz þ y2 )  4(x2  y2  3yz) þ x2 þ y2

(d ) 3  {2x  ½1  (x þ y) þ ½x  2y}

Add the algebraic expressions in each of the following groups. (a) 2x2 þ y2  x þ y,

3y2 þ x  x2 ,

(b) a2  ab þ 2bc þ 3c2 ,

x  2y þ x2  4y2

2ab þ b2  3bc  4c2 ,

ab  4bc þ c2  a2 ,

(c) 2a2 bc  2acb2 þ 5c2 ab, 4b2 ac þ 4bca2  7ac2 b, 2.14

a2 þ 2c2 þ 5bc  2ab

4abc2  3a2 bc  3ab2 c, b2 ac  abc2  3a2 bc

Subtract the second of each of the following expressions from the first. (a) 3xy  2yz þ 4zx, 3zx þ yz  2xy (b) 4x2 þ 3y2  6x þ 4y  2, 2x  y2 þ 3x2  4y þ 3 (c) r 3  3r 2 s þ 4rs2  s3 ,

2.15

2s3 þ 3s2 r  2sr2  3r3

Subtract xy  3yz þ 4xz from twice the sum of the following expressions: 3xy  4yz þ 2xz and 3yz  4zx  2xy.

CHAP. 2]

2.16

Obtain the product of the algebraic expressions in each of the following groups. (a) 4x2 y5 ,

3x3 y2

( f ) y2  4y þ 16,

(b) 3abc2 ,

2a3 b2 c4 , 6a2 b2

(g) x3 þ x2 y þ xy2 þ y3 ,

(c) 4x y, 3xy  4xy 2

(h) x þ 4x þ 8,

2

3

(e) y  4,

(i) 3r  s  t ,

2s þ r þ 3t2

( j ) 3  x  y,

2x þ y þ 1,

2

2r s

yþ3

xy

12x4 yz3 3x2 y4 z

(b)

18r3 s2 t 4r 5 st2

(c)

4ab3  3a2 bc þ 12a3 b2 c4 2ab2 c3

(d )

4x3  5x2 þ 3x  2 xþ1

Perform the indicated divisions. (a)

2.19

2 4

xy

2

Perform the indicated divisions. (a)

2.18

3

yþ4

x  4x þ 8

2

(d ) r s þ 3rs  4rs þ s , 2

2.17

21

FUNDAMENTAL OPERATIONS WITH ALGEBRAIC EXPRESSIONS

27s3  64 3s  4

(b)

1  x2 þ x4 1x

(c)

2y3 þ y5  3y  2 y2  3y þ 1

(d )

4x3 y þ 5x2 y2 þ x4 þ 2xy3 x2 þ 2y2 þ 3xy

Perform the indicated operations and check by using the values x ¼ 1, y ¼ 2. (a) (x4 þ x2 y2 þ y4 )(y4  x2 y2 þ x4 )

(b)

x4 þ xy3 þ x3 y þ 2x2 y2 þ y4 xy þ x2 þ y2

ANSWERS TO SUPPLEMENTARY PROBLEMS 2.10

(a) 24 (b) 212 (c) 1 (d ) 90

2.11

(a) 4

2.12

(a) 3y  x

2.13

(a) 2x2 þ x  y

2.14

(a) 5xy  3yz þ zx

2.15

xy þ yz  8xz

2.16

(a) (b) (c) (d ) (e)

2.17

(a) 

2.18

(a) 9s2 þ 12s þ 16

2.19

(a) x8 þ x4 y4 þ y8 . Check: 21(13) ¼ 273.

(b) 6

(c) 5

(d ) 3

(b) 8y2 þ 6yz

(e) 0

(b) a2 þ b2 þ 2c2

(b)

(f ) 5

(c) 12x  5y

(d ) y  4x þ 4

(c) abc2

(b) x2 þ 4y2  8x þ 8y  5

12x5 y7 36a6 b5 c6 12x3 y3 þ 16x3 y2 2r4 s5 þ 6r 3 s7  8r 3 s5 þ 2r 2 s7 y2  y  12 4x2 z2 y3

(e) 11/5 ( f ) 8 (g) 1=6 (h) 24=5

9s 2r 2 t

(c) 

(f ) (g) (h) (i) (j)

y3 þ 64 x4  y4 x4 þ 64 3r 2 þ 5rs þ 8rt2  2s2  5st2  3t4 y3  2y2  3y þ 3x þ 5x2  3xy  2x3  x2 y þ 2xy2

2b 3a þ  6a2 c c3 2bc2

(b) x3  x2 þ

(c) 4r 3  r 2 s þ rs2  3s3

1 1x

(d ) 4x2  9x þ 12 þ

(c) y3 þ 3y2 þ 10y þ 27 þ (b) x2 þ y2 . Check: 35=7 ¼ 5.

14 xþ1

68y  29 y2  3y þ 1

(d ) x2 þ xy

CHAPTER 3

Properties of Numbers 3.1

SETS OF NUMBERS

The set of counting (or natural) numbers is the set of numbers: 1, 2, 3, 4, 5, . . . . The set of whole numbers is the set of counting numbers and zero: 0, 1, 2, 3, 4, . . . . The set of integers is the set of counting numbers, zero, and the opposites of the counting numbers: . . . , 25, 24, 23, 22, 21, 0, 1, 2, 3, 4, 5, . . . . The set of real numbers is the set of all numbers that correspond to the points on a number line. The real numbers can be separated into two distinct subsets: the rational numbers and the irrational numbers. The set of rational numbers is the set of real numbers that can be written in the form a/b, where a and b are integers and b is not zero. The rational numbers can be thought of as the set of integers and the pffiffiffi common fractions. The numbers 24, 2/3, 50/7, 9, 10/5, 21/2, 0, 145, and 15/1 are examples of rational numbers. setffiffiffi ofpirrational is the set of real numbers that are not rational numbers. The numbers pffiffiffi p pffiffinumbers ffi pffiffiffi The ffiffiffi 2, 3 5, 4 10, 3 þ 4, 3 6  5, and the mathematical constants p and e are examples of irrational numbers.

3.2

PROPERTIES

A set has closure under an operation if the result of performing the operation with two elements of the set is also an element of the set. The set X is closed under the operation  if for all elements a and b in set X, the result a b is in set X. A set has an identity under an operation if there is an element in the set that when combined with each element in the set leaves that element unchanged. The set X has an identity under the operation  if there is an element j in set X such that j a ¼ a j ¼ a for all elements a in set X. A set has inverses under an operation if for each element of the set there is an element of the set such that when these two elements are combined using the operation, the result is the identity for the set under the operation. If a set does not have an identity under an operation, it cannot have the inverse property for the operation. If X is a set that has identity j under operation  , then it has inverses if for each element a in set X there is an element a0 in set X such that a a0 ¼ j and a0  a ¼ j. Sets under an operation may also have the associative property and the commutative property, as described in Section 1.4. If there are two operations on the set then the set could have the distributive property, also described in Section 1.4. EXAMPLE 3.1. Which properties are true for the counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers under the operation of addition?

22

CHAP. 3]

þ

23

PROPERTIES OF NUMBERS

Counting

Whole

Integers

Rational

Irrational

Real

Closure

Yes

Yes

Yes

Yes

No

Yes

Identity

No

Yes

Yes

Yes

No

Yes

Inverse

No

No

Yes

Yes

No

Yes

Associativity

Yes

Yes

Yes

Yes

Yes

Yes

Commutativity

Yes

Yes

Yes

Yes

Yes

Yes

3.3

ADDITIONAL PROPERTIES

There are some properties that sets of numbers have that do not depend on an operation to be true. Three such properties are order, density, and completeness. A set of numbers has order if given two distinct elements in the set one element is greater than the other. A set of numbers has density if between any two elements of the set there is another element of the set. A set of numbers has completeness if the points using its elements as coordinates completely fill a line or plane. EXAMPLE 3.2. Which properties are true for the counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?

Counting

Whole

Integers

Rational

Irrational

Real

Order

Yes

Yes

Yes

Yes

Yes

Yes

Density

No

No

No

Yes

Yes

Yes

Completeness

No

No

No

No

No

Yes

Solved Problems 3.1

Which of the properties closure, identity, and inverse does the set of even integers have under addition? SOLUTION Since even integers are of the form 2n where n is an integer, we let 2m and 2k be any two even integers. The sum of two even integers is 2m þ 2k ¼ 2(m þ k). From Example 3.1 we know that m þ k is an integer, since m and k are integers. Thus, 2(m þ k) is 2 times an integer and is even, so 2m þ 2k is even. Therefore, the even integers are closed under addition. Zero is an even integer since 2(0) ¼ 0. 2m þ 0 ¼ 2m þ 2(0) ¼ 2(m þ 0) ¼ 2m. Thus, 0 is the identity for the even integers under addition. For the even integer 2m, the inverse is 22m. Since m is an integer, 2m is an integer. Thus, 22m ¼ 2(2m) is an even integer. Also, 2m þ (22m) ¼ 2(m þ (2m)) ¼ 2(0) ¼ 0. Therefore, each even integer has an inverse.

3.2

Which of the properties closure, identity, inverse, associativity, and commutativity are true under multiplication for the sets counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?

24

PROPERTIES OF NUMBERS

[CHAP. 3

SOLUTION .

3.3

Counting

Whole

Integers

Rational

Irrational

Real

Closure

Yes

Yes

Yes

Yes

No

Yes

Identity

Yes

Yes

Yes

Yes

No

Yes

Inverse

No

No

No

Yes

No

Yes

Associativity

Yes

Yes

Yes

Yes

Yes

Yes

Commutativity

Yes

Yes

Yes

Yes

Yes

Yes

Which of the properties closure, identity, and inverse does the set of odd integers have under multiplication? SOLUTION Since the odd integers are of the form 2n þ 1 where n is an integer, we let 2m þ 1 and 2k þ 1 be any two odd integers. The product of two odd integers is represented by (2m þ 1)(2k þ 1) ¼ 4mk þ 2m þ 2k þ 1¼ 2(2mk þ m þ k) þ 1. Since the integers are closed under both addition and multiplication, (2mk þ m þ k) is an integer and the product (2m þ 1)(2k þ 1) is equal to 2 times an integer plus 1. Thus, the product is an odd integer. Therefore, the odd integers are closed under multiplication. One is an odd integer, since 2(0) þ 1 ¼ 0 þ 1 ¼ 1. Also (2m þ 1)(1) ¼ (2m)(1) þ (1)(1) ¼ 2m þ 1. Thus, 1 is the identity for the odd integers under multiplication. Seven is an odd integer, since 2(3) þ 1 ¼ 7. Also, 7(1/7) ¼ 1, but 1/7 is not an odd integer. Thus, 7 does not have an inverse under multiplication. Since there is at least one odd integer that does not have an inverse under multiplication, the set of odd integers under multiplication does not have the inverse property.

3.4

Does the set of even integers have the order, density, and completeness properties? SOLUTION Given two distinct even integers 2m and 2k where m and k are integers, we know that either m . k or k . m. If m . k, then 2m . 2k, but if k . m, then 2k . 2m. Thus, the set of even integers has the order property, since for two distinct even integers 2m and 2k either 2m . 2k or 2 k . 2m. The numbers 2m and 2mþ2 are even integers. There is no even integer between 2m and 2m þ 2, since 2m þ 2 ¼ 2(m þ 1) and there is no integer between m and m þ 1. Thus, the even integers do not have the density property. Between the two even integers 8 and 10 is the odd integer 9. Thus, the even integers do not represent the coordinates for all points on a number line. Therefore the even integers do not have the completeness property.

3.5

Let K ¼ f21, 1g. (a) Is K closed under multiplication? (b) Does K have an identity under multiplication? (c) Does K have inverses under multiplication? SOLUTION (a) (1)(1) ¼ 1, (21)(21) ¼ 1, (1)(21) ¼ 21, and (21)(1) ¼ 21. For all possible products of two elements in K, the result is in K. Thus, K is closed under multiplication. (b) 1 is in K, (1)(1) ¼ 1, and (1)(21) ¼ 21. Thus 1 is the identity for K under multiplication. (c) Since (1)(1) ¼ 1 and (21)(21) ¼ 1, each element of K is its own inverse.

CHAP. 3]

25

PROPERTIES OF NUMBERS

Supplementary Problems 3.6

Which of the properties closure, identity, and inverse does the set of even integers have under multiplication?

3.7

Which of the properties closure, identity, and inverse does the set of odd integers have under addition?

3.8

Does the set of odd integers have the order, density, and completeness properties?

3.9

Which of the properties closure, identity, inverse, associativity, and commutativity are true under subtraction for the sets of counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?

3.10

Which of the properties closure, identity, inverse, associativity, and commutativity are true under non-zero division for the set of counting numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers?

3.11

Which of the properties closure, identity, inverse, associativity, and commutativity are true for the set of zero, f0g, under (a) addition, (b) subtraction, and (c) multiplication?

3.12

Which of the properties closure, identity, inverse, associativity, and commutativity are true for the set of one, f1g, under (a) addition, (b) subtraction, (c) multiplication, and (d) division?

ANSWERS TO SUPPLEMENTARY PROBLEMS 3.6

Closure: yes; identity: no; inverse: no.

3.7

Closure: no; identity: no; inverse: no.

3.8

Order: yes; density: no; completeness: no.

3.9

3.10

2

Counting

Whole

Integers

Rational

Irrational

Real

Closure

No

No

Yes

Yes

No

Yes

Identity

No

No

No

No

No

No

Inverse

No

No

No

No

No

No

Associativity

No

No

No

No

No

No

Commutativity

No

No

No

No

No

No

4

Counting

Whole

Integers

Rational

Irrational

Real

Closure

No

No

No

Yes

No

Yes

Identity

No

No

No

No

No

No

Inverse

No

No

No

No

No

No

Associativity

No

No

No

No

No

No

Commutativity

No

No

No

No

No

No

26

PROPERTIES OF NUMBERS

3.11

[CHAP. 3

Closure

Identity

Inverse

Associativity

Commutativity

(a)

þ

Yes

Yes

Yes

Yes

Yes

(b)

2

Yes

Yes

Yes

Yes

Yes

(c)



Yes

Yes

Yes

Yes

Yes

Closure

Identity

Inverse

Associativity

Commutativity

3.12 (a)

þ

No

No

No

Yes

Yes

(b)

2

No

No

No

No

Yes

(c)



Yes

Yes

Yes

Yes

Yes

(d)

4

Yes

Yes

Yes

Yes

Yes

CHAPTER 4

Special Products 4.1

SPECIAL PRODUCTS

The following are some of the products which occur frequently in mathematics, and the student should become familiar with them as soon as possible. Proofs of these results may be obtained by performing the multiplications. I. Product of a monomial and a binomial a(c þ d) ¼ ac þ ad II. Product of the sum and the difference of two terms (a þ b)(a  b) ¼ a2  b2 III. Square of a binomial (a þ b)2 ¼ a2 þ 2ab þ b2 (a  b)2 ¼ a2  2ab þ b2 IV. Product of two binomials (x þ a)(x þ b) ¼ x2 þ (a þ b)x þ ab (ax þ b)(cx þ d) ¼ acx2 þ (ad þ bc)x þ bd (a þ b)(c þ d) ¼ ac þ bc þ ad þ bd V. Cube of a binomial (a þ b)3 ¼ a3 þ 3a2 b þ 3ab2 þ b3 (a  b)3 ¼ a3  3a2 b þ 3ab2  b3 VI. Square of a trinomial (a þ b þ c)2 ¼ a2 þ b2 þ c2 þ 2ab þ 2ac þ 2bc 27

28

4.2

SPECIAL PRODUCTS

[CHAP. 4

PRODUCTS YIELDING ANSWERS OF THE FORM an + bn

It may be verified by multiplication that (a  b)(a2 þ ab þ b2 ) ¼ a3  b3 (a  b)(a3 þ a2 b þ ab2 þ b3 ) ¼ a4  b4 (a  b)(a4 þ a3 b þ a2 b2 þ ab3 þ b4 ) ¼ a5  b5 (a  b)(a5 þ a4 b þ a3 b2 þ a2 b3 þ ab4 þ b5 ) ¼ a6  b6 etc., the rule being clear. These may be summarized by VII.

(a  b)(an1 þ an2 b þ an3 b2 þ    þ abn2 þ bn1 ) ¼ an  bn

where n is any positive integer (1, 2, 3, 4, . . . ). Similarly, it may be verified that (a þ b)(a2  ab þ b2 ) ¼ a3 þ b3 (a þ b)(a4  a3 b þ a2 b2  ab3 þ b4 ) ¼ a5 þ b5 (a þ b)(a6  a5 b þ a4 b2  a3 b3 þ a2 b4  ab5 þ b6 ) ¼ a7 þ b7 etc., the rule being clear. These may be summarized by VIII.

(a þ b)(an1  an2 b þ an3 b2      abn2 þ bn1 ) ¼ an þ bn

where n is any positive odd integer (1, 3, 5, 7, . . . ).

Solved Problems Find each of the following products. 4.1

(a) 3x(2x þ 3y) ¼ (3x)(2x) þ (3x)(3y) ¼ 6x2 þ 9xy, using I with a ¼ 3x, c ¼ 2x, d ¼ 3y: (b) x2 y(3x3  2y þ 4) ¼ (x2 y)(3x3 ) þ (x2 y)(2y) þ (x2 y)(4) ¼ 3x5 y  2x2 y2 þ 4x2 y (c) (3x3 y2 þ 2xy  5)(x2 y3 ) ¼ (3x3 y2 )(x2 y3 ) þ (2xy)(x2 y3 ) þ (5)(x2 y3 ) ¼ 3x5 y5 þ 2x3 y4  5x2 y3 (d ) (2x þ 3y)(2x  3y) ¼ (2x)2  (3y)2 ¼ 4x2  9y2 , using II with a ¼ 2x, b ¼ 3y. (e) (1  5x3 )(1 þ 5x3 ) ¼ (1)2  (5x3 )2 ¼ 1  25x6 ( f ) (5x þ x3 y2 )(5x  x3 y2 ) ¼ (5x)2  (x3 y2 )2 ¼ 25x2  x6 y4 (g) (3x þ 5y)2 ¼ (3x)2 þ 2(3x)(5y) þ (5y)2 ¼ 9x2 þ 30xy þ 25y2 , using III with a ¼ 3x, b ¼ 5y: (h) (x þ 2)2 ¼ x2 þ 2(x)(2) þ 22 ¼ x2 þ 4x þ 4 (i) (7x2  2xy)2 ¼ (7x2 )2  2(7x2 )(2xy) þ (2xy)2 ¼ 49x4  28x3 y þ 4x2 y2 , using III with a ¼ 7x2 , b ¼ 2xy. ( j ) (ax  2by)2 ¼ (ax)2  2(ax)(2by) þ (2by)2 ¼ a2 x2  4axby þ 4b2 y2 (k ) (x4 þ 6)2 ¼ (x4 )2 þ 2(x4 )(6) þ (6)2 ¼ x8 þ 12x4 þ 36 (l) (3y2  2)2 ¼ (3y2 )2  2(3y2 )(2) þ (2)2 ¼ 9y4  12y2 þ 4

CHAP. 4]

SPECIAL PRODUCTS

(m) (x þ 3)(x þ 5) ¼ x2 þ (3 þ 5)x þ (3)(5) ¼ x2 þ 8x þ 15, using IV with a ¼ 3, b ¼ 5. (n) (x  2)(x þ 8) ¼ x2 þ (2 þ 8)x þ (2)(8) ¼ x2 þ 6x  16 (o) (x þ 2)(x  8) ¼ x2 þ (2  8)x þ (2)(8) ¼ x2  6x  16 ( p) (t2 þ 10)(t2  12) ¼ (t2 )2 þ (10  12)t2 þ (10)(12) ¼ t4  2t2  120 (q) (3x þ 4)(2x  3) ¼ (3)(2)x2 þ ½(3)(3) þ (4)(2)x þ (4)(3) ¼ 6x2  x  12, using IV with a ¼ 3, b ¼ 4, c ¼ 2, d ¼ 3. (r) (2x þ 5)(4x  1) ¼ (2)(4)x2 þ ½(2)(1) þ (5)(4)x þ (5)(1) ¼ 8x2 þ 18x  5 (s) (3x þ y)(4x  2y) ¼ (3x)(4x) þ (y)(4x) þ (3x)(2y) þ (y)(2y) ¼ 12x2  2xy  2y2 , using V with a ¼ 3x, b ¼ y, c ¼ 4x, d ¼ 2y. (t) (3t2 s  2)(4t  3s) ¼ (3t2 s)(4t) þ (2)(4t) þ (3t2 s)(3s) þ (2)(3s) ¼ 12t3 s  8t  9t2 s2 þ 6s (u) (3xy þ 1)(2x2  3y) ¼ (3xy)(2x2 ) þ (3xy)(3y) þ (1)(2x2 ) þ (1)(3y) ¼ 6x3 y  9xy2 þ 2x2  3y (v) (x þ y þ 3)(x þ y  3) ¼ (x þ y)2  32 ¼ x2 þ 2xy þ y2  9 (w) (2x  y  1)(2x  y þ 1) ¼ (2x  y)2  (1)2 ¼ 4x2  4xy þ y2  1 (x) (x2 þ 2xy þ y2 )(x2  2xy þ y2 ) ¼ (x2 þ y2 þ 2xy)(x2 þ y2  2xy) ¼ (x2 þ y2 )2  (2xy)2 ¼ x4 þ 2x2 y2 þ y4  4x2 y2 ¼ x4  2x2 y2 þ y4 (y) (x3 þ 2 þ xy)(x3  2 þ xy) ¼ (x3 þ xy þ 2)(x3 þ xy  2) ¼ (x3 þ xy)2  22 ¼ x6 þ 2(x3 )(xy) þ (xy)2  4 ¼ x6 þ 2x4 y þ x2 y2  4 4.2

(a) (x þ 2y)3 ¼ x3 þ 3(x)2 (2y) þ 3(x)(2y)2 þ (2y)3 ¼ x3 þ 6x2 y þ 12xy2 þ 8y3 , using V with a ¼ x, b ¼ 2y. (b) (3x þ 2)3 ¼ (3x)3 þ 3(3x)2 (2) þ 3(3x)(2)2 þ (2)3 ¼ 27x3 þ 54x2 þ 36x þ 8 (c) (2y  5)3 ¼ (2y)3  3(2y)2 (5) þ 3(2y)(5)2  (5)3 ¼ 8y3  60y2 þ 150y  125; using V with a ¼ 2y, b ¼ 5. (d ) (xy  2)3 ¼ (xy)3  3(xy)2 (2) þ 3(xy)(2)2  (2)3 ¼ x3 y3  6x2 y2 þ 12xy  8 (e) (x2 y  y2 )3 ¼ (x2 y)3  3(x2 y)2 (y2 ) þ 3(x2 y)(y2 )2  (y2 )3 ¼ x6 y3  3x4 y4 þ 3x2 y5  y6 ( f ) (x  1)(x2 þ x þ 1) ¼ x3  1, using VII with a ¼ x, b ¼ 1. If the form is not recognized, multiply as follows. (x  1)(x2 þ x þ 1) ¼ x(x2 þ x þ 1)  1(x2 þ x þ 1) ¼ x3 þ x2 þ x  x2  x  1 ¼ x3  1 (g) (x  2y)(x2 þ 2xy þ 4y2 ) ¼ x3  (2y)3 ¼ x3  8y3 , using VII with a ¼ x, b ¼ 2y. (h) (xy þ 2)(x2 y2  2xy þ 4) ¼ (xy)3 þ (2)3 ¼ x3 y3 þ 8, using VIII with a ¼ xy, b ¼ 2. (i) (2x þ 1)(4x2  2x þ 1) ¼ (2x)3 þ 1 ¼ 8x3 þ 1 ( j ) (2x þ 3y þ z)2 ¼ (2x)2 þ (3y)2 þ (z)2 þ 2(2x)(3y) þ 2(2x)(z) þ 2(3y)(z) ¼ 4x2 þ 9y2 þ z2 þ 12xy þ 4xz þ 6yz, using VI with a ¼ 2x, b ¼ 3y, c ¼ z. (k ) (u3  v2 þ 2w)2 ¼ (u3 )2 þ (v2 )2 þ (2w)2 þ 2(u3 )(v2 ) þ 2(u3 )(2w) þ 2(v2 )(2w) ¼ u6 þ v4 þ 4w2  2u3 v2 þ 4u3 w  4v2 w

4.3

(a) (x  1)(x5 þ x4 þ x3 þ x2 þ x þ 1) ¼ x6  1, using VII with a ¼ x, b ¼ 1, n ¼ 6. (b) (x  2y)(x4 þ 2x3 y þ 4x2 y2 þ 8xy3 þ 16y4 ) ¼ x5  (2y)5 ¼ x5  32y5 , using VII with a ¼ x, b ¼ 2y.

29

30

SPECIAL PRODUCTS

[CHAP. 4

(c) (3y þ x)(81y4  27y3 x þ 9y2 x2  3yx3 þ x4 ) ¼ (3y)5 þ x5 ¼ 243y5 þ x5 ; using VIII with a ¼ 3y, b ¼ x.

4.4

(a) (x þ y þ z)(x þ y  z)(x  y þ z)(x  y  z). The first two factors may be written as (x þ y þ z)(x þ y  z) ¼ (x þ y)2  z2 ¼ x2 þ 2xy þ y2  z2 , and the second two factors as (x  y þ z)(x  y  z) ¼ (x  y)2  z2 ¼ x2  2xy þ y2  z2 : The result may be written (x2 þ y2  z2 þ 2xy)(x2 þ y2  z2  2xy) ¼ (x2 þ y2  z2 )2  (2xy)2 ¼ (x2 )2 þ ðy2 Þ2 þ ðz2 Þ2 þ 2ðx2 Þðy2 Þ þ 2(x2 Þðz2 Þ þ 2ðy2 Þ(z2 )  4x2 y2 ¼ x4 þ y4 þ z4 þ 2x2 y2  2x2 z2  2y2 z2 (b) (x þ y þ z þ 1)2 ¼ ½(x þ y) þ (z þ 1)2 ¼ (x þ y)2 þ 2(x þ y)(z þ 1) þ (z þ 1)2 ¼ x2 þ 2xy þ y2 þ 2xz þ 2x þ 2yz þ 2y þ z2 þ 2z þ 1 (c) (u  v)3 (u þ v)3 ¼ ½(u  v)(u þ v)3 ¼ (u2  v2 )3 ¼ (u2 )3  3(u2 )2 v2 þ 3(u2 )(v2 )2  (v2 )3 ¼ u6  3u4 v2 þ 3u2 v4  v6 (d ) (x2  x þ 1)2 (x2 þ x þ 1)2 ¼ ½(x2  x þ 1)(x2 þ x þ 1)2 ¼ ½(x2 þ 1  x)(x2 þ 1 þ x)2 ¼ ½(x2 þ 1)2  x2 2 ¼ ½x4 þ 2x2 þ 1  x2 2 ¼ (x4 þ x2 þ 1)2 ¼ (x4 )2 þ (x2 )2 þ 12 þ 2(x4 )(x2 ) þ 2(x4 )(1) þ 2(x2 )(1) ¼ x8 þ x4 þ 1 þ 2x6 þ 2x4 þ 2x2 ¼ x8 þ 2x6 þ 3x4 þ 2x2 þ 1 (e) (ey þ 1)(ey  1)(e2y þ 1)(e4y þ 1)(e8y þ 1) ¼ (e2y  1)(e2y þ 1)(e4y þ 1)(e8y þ 1) ¼ (e4y  1)(e4y þ 1)(e8y þ 1) ¼ (e8y  1)(e8y þ 1) ¼ e16y  1

Supplementary Problems Find each of the following products. 4.5

(a) 2xy(3x2 y  4y3 ) ¼ 6x3 y2  8xy4 (b) 3x2 y3 (2xy  x  2y) ¼ 6x3 y4  3x3 y3  6x2 y4 (c) (2st3  4rs2 þ 3s3 t)(5rst2 ) ¼ 10rs2 t5  20r 2 s3 t2 þ 15rs4 t3 (d ) (3a þ 5b)(3a  5b) ¼ 9a2  25b2 (e) (5xy þ 4)(5xy  4) ¼ 25x2 y2  16 ( f ) (2  5y2 )(2 þ 5y2 ) ¼ 4  25y4 (g) (3a þ 5a2 b)(3a  5a2 b) ¼ 9a2  25a4 b2 (h) (x þ 6)2 ¼ x2 þ 12x þ 36 (i) (y þ 3x)2 ¼ y2 þ 6xy þ 9x2 ( j ) (z  4)2 ¼ z2  8z þ 16 (k ) (3  2x2 )2 ¼ 9  12x2 þ 4x4 (l) (x2 y  2z)2 ¼ x4 y2  4x2 yz þ 4z2 (m) (x þ 2)(x þ 4) ¼ x2 þ 6x þ 8

CHAP. 4]

SPECIAL PRODUCTS

(n) (x  4)(x þ 7) ¼ x2 þ 3x  28 (o) (y þ 3)(y  5) ¼ y2  2y  15 ( p) (xy þ 6)(xy  4) ¼ x2 y2 þ 2xy  24 (q) (2x  3)(4x þ 1) ¼ 8x2  10x  3 (r) (4 þ 3r)(2  r) ¼ 8 þ 2r  3r 2 (s) (5x þ 3y)(2x  3y) ¼ 10x2  9xy  9y2 (t) (2t2 þ s)(3t2 þ 4s) ¼ 6t4 þ 11t2 s þ 4s2 (u) (x2 þ 4y)(2x2 y  y2 ) ¼ 2x4 y þ 7x2 y2  4y3 (v) x(2x  3)(3x þ 4) ¼ 6x3  x2  12x (w) (r þ s  1)(r þ s þ 1) ¼ r 2 þ 2rs þ s2  1 (x) (x  2y þ z)(x  2y  z) ¼ x2  4xy þ 4y2  z2 (y) (x2 þ 2x þ 4)(x2  2x þ 4) ¼ x4 þ 4x2 þ 16

4.6

(a) (2x þ 1)3 ¼ 8x3 þ 12x2 þ 6x þ 1 (b) (3x þ 2y)3 ¼ 27x3 þ 54x2 y þ 36xy2 þ 8y3 (c) (r  2s)3 ¼ r 3  6r 2 s þ 12rs2  8s3 (d ) (x2  1)3 ¼ x6  3x4 þ 3x2  1 (e) (ab2  2b)3 ¼ a3 b6  6a2 b5 þ 12ab4  8b3 (f ) (t  2)(t2 þ 2t þ 4) ¼ t3  8 (g) (z  x)(x2 þ xz þ z2 ) ¼ z3  x3 (h) (x þ 3y)(x2  3xy þ 9y2 ) ¼ x3 þ 27y3

4.7

(a) (x  2y þ z)2 ¼ x2  4xy þ 4y2 þ 2zx  4zy þ z2 (b) (s  1)(s3 þ s2 þ s þ 1) ¼ s4  1 (c) (1 þ t2 )(1  t2 þ t4  t6 ) ¼ 1  t8 (d ) (3x þ 2y)2 (3x  2y)2 ¼ 81x4  72x2 y2 þ 16y4 (e) (x2 þ 2x þ 1)2 (x2  2x þ 1)2 ¼ x8  4x6 þ 6x4  4x2 þ 1 ( f ) (y  1)3 (y þ 1)3 ¼ y6  3y4 þ 3y2  1 (g) (u þ 2)(u  2)(u2 þ 4)(u4 þ 16) ¼ u8  256

31

CHAPTER 5

Factoring 5.1 FACTORING The factors of a given algebraic expression consist of two or more algebraic expressions which when multiplied together produce the given expression. EXAMPLES 5.1.

Factor each algebraic expression.

(a) x2  7x þ 6 ¼ (x  1)(x  6) (b) x2 þ 8x ¼ x(x þ 8) (c) 6x2  7x  5 ¼ (3x  5)(2x þ 1) (d ) x2 þ 2xy  8y2 ¼ (x þ 4y)(x  2y) The factorization process is generally restricted to finding factors of polynomials with integer coefficients in each of its terms. In such cases it is required that the factors also be polynomials with integer coefficients. Unless otherwise stated we shall adhere to this limitation. pffiffiffi pffiffiffi Thus we shall not consider (x  1) as being factorable into ( x þ 1)( x  1) because these pffiffiffi factors pffiffiare ffi not polynomials. Similarly, we shall not consider (x2  3y2 ) as being factorable into (x  3y)(x þ 3yÞ because these factors are not polynomials with integer coefficients. Also, even though 3x þ 2y could be written 3(x þ 23 y) we shall not consider this to be a factored form because x þ 23 y is not a polynomial with integer coefficients. A given polynomial with integer coefficients is said to be prime if it cannot itself be factored in accordance with the above restrictions. Thus x2  7x þ 6 ¼ (x  1)(x  6) has been expressed as a product of the prime factors x  1 and x  6. A polynomial is said to be factored completely when it is expressed as a product of prime factors. Note 1. In factoring we shall allow trivial changes in sign. Thus x2  7x þ 6 can be factored either as (x  1)(x  6) or (1  x)(6  x). It can be shown that factorization into prime factors, apart from the trivial changes in sign and arrangement of factors, is possible in one and only one way. This is often referred to as the Unique Factorization Theorem. Note 2. Sometimes the following definition of prime is used. A polynomial is said to be prime if it has no factors other than plus or minus itself and +1. This is in analogy with the definition of a prime number or integer such as 2, 3, 5, 7, 11, . . . and may be seen to be equivalent to the previous definition. Note 3. Occasionally we may factor polynomials with rational coefficients, e.g., x2  9=4 ¼ (x þ 3=2)(x  3=2). In such cases the factors should be polynomials with rational coefficients. 32

CHAP. 5]

FACTORING

33

2 Note pffiffiffi 4. There pffiffiffi are times when we want to factor an expression over a specific set of numbers, e.g., x  2 ¼ (x þ 2)(x  2) over the set of real numbers, but it is prime over the set of rational numbers. Unless the set of numbers to use for the coefficients of the factors is specified it is assumed to be the set of integers.

5.2 FACTORIZATION PROCEDURES In factoring, formulas I-VIII of Chapter 4 are very useful. Just as when read from left to right they helped to obtain products, so when read from right to left they help to find factors. The following procedures in factoring are very useful. A.

Common monomial factor. Type: ac þ ad ¼ a(c þ d) EXAMPLES 5.2. (a) 6x2 y  2x3 ¼ 2x2 (3y  x) (b) 2x3 y  xy2 þ 3x2 y ¼ xy(2x2  y þ 3x)

B.

Difference of two squares. Type: a2  b2 ¼ (a þ b)(a  b) EXAMPLES 5.3. (a) x2  25 ¼ x2  52 ¼ (x þ 5)(x  5) where a ¼ x, b ¼ 5 (b) 4x2  9y2 ¼ (2x)2  (3y)2 ¼ (2x þ 3y)(2x  3y) where a ¼ 2x, b ¼ 3y

C.

Perfect square trinomials. Types: a2 þ 2ab þ b2 ¼ (a þ b)2 a2  2ab þ b2 ¼ (a  b)2

It follows that a trinomial is a perfect square if two terms are perfect squares and the third term is numerically twice the product of the square roots of the other two terms. EXAMPLES 5.4. (a) x2 þ 6x þ 9 ¼ (x þ 3)2 (b) 9x2  12xy þ 4y2 ¼ (3x  2y)2

D.

Other trinomials. Types: x2 þ (a þ b)x þ ab ¼ (x þ a)(x þ b) acx2 þ (ad þ bc)x þ bd ¼ (ax þ b)(cx þ d) EXAMPLES 5.5.

E.

(a) x2  5x þ 4 ¼ (x  4)(x  1) where a ¼ 4, b ¼ 1 so that their sum (a þ b) ¼ 5 and their product ab ¼ 4. (b) x2 þ xy  12y2 ¼ (x  3y)(x þ 4y) where a ¼ 3y, b ¼ 4y (c) 3x2  5x  2 ¼ (x  2)(3x þ 1): Here ac ¼ 3, bd ¼ 2, ad þ bc ¼ 5; and we find by trial that a ¼ 1, c ¼ 3, b ¼ 2, d ¼ 1 satisfies ad þ bc ¼ 5. (d) 6x2 þ x  12 ¼ (3x  4)(2x þ 3) (e) 8  14x þ 5x2 ¼ (4  5x)(2  x)

Sum, difference of two cubes. Types: a3 þ b3 ¼ (a þ b)(a2  ab þ b2 ) a3  b3 ¼ (a  b)(a2 þ ab þ b2 ) EXAMPLES 5.6. (a) 8x3 þ 27y3 ¼ (2x)3 þ (3y)3 ¼ (2x þ 3y)½(2x)2  (2x)(3y) þ (3y)2  ¼ (2x þ 3y)(4x2  6xy þ 9y2 ) 3 3 (b) 8x y  1 ¼ (2xy)3  13 ¼ (2xy  1Þ(4x2 y2 þ 2xy þ 1)

F.

Grouping of terms. Type: ac þ bc þ ad þ bd ¼ c(a þ b) þ d(a þ b) ¼ (a þ b)(c þ d) EXAMPLE 5.7.

G.

2ax  4bx þ ay  2by ¼ 2x(a  2b) þ y(a  2b) ¼ (a  2b)(2x þ y)

Factors of an + bn . Here we use formulas VII and VIII of Chapter 4.

34

FACTORING

EXAMPLES 5.8.

H.

[CHAP. 5

(a) 32x5 þ 1 ¼ (2x)5 þ 15 ¼ (2x þ 1)½(2x)4  (2x)3 þ (2x)2  2x þ 1 ¼ (2x þ 1)(16x4  8x3 þ 4x2  2x þ 1) (b) x7  1 ¼ (x  1)(x6 þ x5 þ x4 þ x3 þ x2 þ x þ 1)

Addition and subtraction of suitable terms. EXAMPLE 5.9.

Factor x4 þ 4. Adding and subtracting 4x2 (twice the product of the square roots of x4 and 4), we have x4 þ 4 ¼ (x4 þ 4x2 þ 4)  4x2 ¼ (x2 þ 2)2  (2x)2 ¼ (x2 þ 2 þ 2x)(x2 þ 2  2x) ¼ (x2 þ 2x þ 2)(x2  2x þ 2)

I.

Miscellaneous combinations of previous methods. EXAMPLES 5.10.

(a) x4  xy3  x3 y þ y4 ¼ (x4  xy3 )  (x3 y  y4 ) ¼ x(x3  y3 )  y(x3  y3 ) ¼ (x3  y3 )(x  y) ¼ (x  y)(x2 þ xy þ y2 )(x  y) ¼ (x  y)2 (x2 þ xy þ y2 ) (b) x2 y  3x2  y þ 3 ¼ (x2 y  3x2 ) þ (y þ 3) ¼ x2 (y  3)  (y  3) ¼ (y  3)(x2  1) ¼ (y  3)(x þ 1)(x  1) (c) x2 þ 6x þ 9  y2 ¼ (x2 þ 6x þ 9)  y2 ¼ (x þ 3)2  y2 ¼ ½(x þ 3) þ y½(x þ 3)  y ¼ (x þ y þ 3)(x  y þ 3)

5.3 GREATEST COMMON FACTOR The greatest common factor (GCF) of two or more given polynomials is the polynomial of highest degree and largest numerical coefficients (apart from trivial changes in sign) which is a factor of all the given polynomials. The following method is suggested for finding the GCF of several polynomials. (a) Write each polynomial as a product of prime factors. (b) The GCF is the product obtained by taking each factor to the lowest power to which it occurs in any of the polynomials. EXAMPLE 5.11.

The GCF of 23 32 (x  y)3 (x þ 2y)2 , 22 33 (x  y)2 (x þ 2y)3 , 32 (x  y)2 (x þ 2y) is 32 (x  y)2 (x þ 2y).

Two or more polynomials are relatively prime if their GCF is 1.

5.4 LEAST COMMON MULTIPLE The least common multiple (LCM) of two or more given polynomials is the polynomial of lowest degree and smallest numerical coefficients (apart from trivial changes in sign) for which each of the given polynomials will be a factor. The following procedure is suggested for determining the LCM of several polynomials. (a) Write each polynomial as a product of prime factors. (b) The LCM is the product obtained by taking each factor to the highest power to which it occurs. EXAMPLE 5.12.

The LCM of 23 32 (x  y)3 (x þ 2y)2 , 22 33 (x  y)2 (x þ 2y)3 , 32 (x  y)2 (x þ 2y) is 23 33 (x  y)3 (x þ 2y)3 :

CHAP. 5]

FACTORING

35

Solved Problems Common Monomial Factor Type: ac þ ad ¼ a(c þ d) 5.1

(a) (b) (c) (d) (e) (f)

2x2  3xy ¼ x(2x  3y) 4x þ 8y þ 12z ¼ 4(x þ 2y þ 3z) 3x2 þ 6x3 þ 12x4 ¼ 3x2 (1 þ 2x þ 4x2 ) 9s3 t þ 15s2 t3  3s2 t2 ¼ 3s2 t(3s þ 5t2  t) 10a2 b3 c4  15a3 b2 c4 þ 30a4 b3 c2 ¼ 5a2 b2 c2 (2bc2  3ac2 þ 6a2 b) 4anþ1  8a2n ¼ 4anþ1 (1  2an1 )

Difference of Two Squares Type: a2  b2 ¼ (a þ b)(a  b) 5.2

(a) x2  9 ¼ x2  32 ¼ (x þ 3)(x  3) (b) 25x2  4y2 ¼ (5x)2  (2y)2 ¼ (5x þ 2y)(5x  2y) (c) 9x2 y2  16a2 ¼ (3xy)2  (4a)2 ¼ (3xy þ 4a)(3xy  4a) (d) 1  m2 n4 ¼ 12  (mn2 )2 ¼ (1 þ mn2 )(1  mn2 ) (e) 3x2  12 ¼ 3(x2  4) ¼ 3(x þ 2)(x  2) ( f ) x2 y2  36y4 ¼ y2 ½x2  (6y)2  ¼ y2 (x þ 6y)(x  6y) (g) x4  y4 ¼ (x2 )2  (y2 )2 ¼ (x2 þ y2 )(x2  y2 ) ¼ (x2 þ y2 )(x þ y)(x  y) (h) 1  x8 ¼ (1 þ x4 )(1  x4 ) ¼ (1 þ x4 )(1 þ x2 )(1  x2 ) ¼ (1 þ x4 )(1 þ x2 )(1 þ x)(1  x) (i) 32a4 b  162b5 ¼ 2b(16a4  81b4 ) ¼ 2b(4a2 þ 9b2 )(4a2  9b2 ) ¼ 2b(4a2 þ 9b2 )(2a þ 3b)(2a  3b) ( j) x3 y  y3 x ¼ xy(x2  y2 ) ¼ xy(x þ y)(x  y) (k) (x þ 1)2  36y2 ¼ ½(x þ 1) þ (6y)½(x þ 1)  (6y) ¼ (x þ 6y þ 1)(x  6y þ 1) (l ) (5x þ 2y)2  (3x  7y)2 ¼ ½(5x þ 2y) þ (3x  7y)½(5x þ 2y)  (3x  7y) ¼ (8x  5y)(2x þ 9y) Perfect Square Trinomials Types: a2 þ 2ab þ b2 ¼ (a þ b)2 a2  2ab þ b2 ¼ (a  b)2

5.3

(a) (b) (c) (d) (e) (f) (g) (h) (i) ( j) (k) (l )

x2 þ 8x þ 16 ¼ x2 þ 2(x)(4) þ 42 ¼ (x þ 4)2 1 þ 4y þ 4y2 ¼ (1 þ 2y)2 t2  4t þ 4 ¼ t2  2(t)(2) þ 22 ¼ (t  2)2 x2  16xy þ 64y2 ¼ (x  8y)2 25x2 þ 60xy þ 36y2 ¼ (5x þ 6y)2 16m2  40mn þ 25n2 ¼ (4m  5n)2 9x4  24x2 y þ 16y2 ¼ (3x2  4y)2 2x3 y3 þ 16x2 y4 þ 32xy5 ¼ 2xy3 (x2 þ 8xy þ 16y2 ) ¼ 2xy3 (x þ 4y)2 16a4  72a2 b2 þ 81b4 ¼ (4a2  9b2 )2 ¼ ½(2a þ 3b)(2a  3b)2 ¼ (2a þ 3b)2 (2a  3b)2 (x þ 2y)2 þ 10(x þ 2y) þ 25 ¼ (x þ 2y þ 5)2 a2 x2  2abxy þ b2 y2 ¼ (ax  by)2 4m6 n6 þ 32m4 n4 þ 64m2 n2 ¼ 4m2 n2 (m4 n4 þ 8m2 n2 þ 16) ¼ 4m2 n2 (m2 n2 þ 4)2

Other Trinomials Types: x2 þ (a þ b)x þ ab ¼ (x þ a)(x þ b) acx2 þ (ad þ bc)x þ bd ¼ (ax þ b)(cx þ d)

36

5.4

FACTORING

(a) (b) (c) (d ) (e) (f ) (g) (h) (i) (j) (k) (l ) (m) (n) (o) (p) (q) (r) (s) (t)

5.5

[CHAP. 5

x2 þ 6x þ 8 ¼ (x þ 4)(x þ 2) x2  6x þ 8 ¼ (x  4)(x  2) x2 þ 2x  8 ¼ (x þ 4)(x  2) x2  2x  8 ¼ (x  4)(x þ 2) x2  7xy þ 12y2 ¼ (x  3y)(x  4y) x2 þ xy  12y2 ¼ (x þ 4y)(x  3y) 16  10x þ x2 ¼ (8  x)(2  x) 20  x  x2 ¼ (5 þ x)(4  x) 3x3  3x2  18x ¼ 3x(x2  x  6) ¼ 3x(x  3)(x þ 2) y4 þ 7y2 þ 12 ¼ (y2 þ 4)(y2 þ 3) m4 þ m2  2 ¼ (m2 þ 2)(m2  1) ¼ (m2 þ 2)(m þ 1)(m  1) (x þ 1)2 þ 3(x þ 1) þ 2 ¼ ½(x þ 1) þ 2½(x þ 1) þ 1 ¼ (x þ 3)(x þ 2) s2 t2  2st3  63t4 ¼ t2 (s2  2st  63t2 ) ¼ t2 (s  9t)(s þ 7t) z4  10z2 þ 9 ¼ (z2  1)(z2  9) ¼ (z þ 1)(z  1)(z þ 3)(z  3) 2x6 y6x4 y3 8x2 y5 ¼ 2x2 y(x4 3x2 y2 4y4 ) ¼ 2x2 y(x2 þy2 )(x2 4y2 ) ¼ 2x2 y(x2 þy2 )(xþ2y)(x2y) x2  2xy þ y2 þ 10(x  y) þ 9 ¼ (x  y)2 þ 10(x  y) þ 9 ¼ ½(x  y) þ 1½(x  y) þ 9 ¼ (x  y þ 1)(x  y þ 9) 4x8 y10  40x5 y7 þ 84x2 y4 ¼ 4x2 y4 (x6 y6  10x3 y3 þ 21) ¼ 4x2 y4 (x3 y3  7)(x3 y3  3) x2a  xa  30 ¼ (xa  6)(xa þ 5) xmþ2n þ 7xmþn þ 10xm ¼ xm (x2n þ 7xn þ 10) ¼ xm (xn þ 2)(xn þ 5) a2(y1)  5ay1 þ 6 ¼ (ay1  3)(ay1  2)

3x2 þ 10x þ 3 ¼ (3x þ 1)(x þ 3) 2x2  7x þ 3 ¼ (2x  1)(x  3) 2y2  y  6 ¼ (2y þ 3)(y  2) 10s2 þ 11s  6 ¼ (5s  2)(2s þ 3) 6x2  xy  12y2 ¼ (3x þ 4y)(2x  3y) 10  x  3x2 ¼ (5  3x)(2 þ x) 4z4  9z2 þ 2 ¼ (z2  2)(4z2  1) ¼ (z2  2)(2z þ 1)(2z  1) 16x3 y þ 28x2 y2  30xy3 ¼ 2xy(8x2 þ 14xy  15y2 ) ¼ 2xy(4x  3y)(2x þ 5y) 12(x þ y)2 þ 8(x þ y)  15 ¼ ½6(x þ y)  5½2(x þ y) þ 3 ¼ (6x þ 6y  5)(2x þ 2y þ 3) 6b2nþ1 þ 5bnþ1  6b ¼ b(6b2n þ 5bn  6) ¼ b(2bn þ 3)(3bn  2) 18x4pþm  66x2pþm y2  24xm y4 ¼ 6xm (3x4p  11x2p y2  4y4 ) ¼ 6xm (3x2p þ y2 )(x2p  4y2 ) ¼ 6xm (3x2p þ y2 )(xp þ 2y)(xp  2y) 12 3 8 7 4 11 (l) 64x y  68x y þ 4x y ¼ 4x4 y3 (16x8  17x4 y4 þ y8 ) ¼ 4x4 y3 (16x4  y4 )(x4  y4 ) ¼ 4x4 y3 (4x2 þ y2 )(4x2  y2 )(x2 þ y2 )(x2  y2 ) ¼ 4x4 y3 (4x2 þ y2 )(2x þ y)(2x  y)(x2 þ y2 )(x þ y)(x  y)

(a) (b) (c) (d ) (e) (f) (g) (h) (i) (j) (k)

Sum of Difference of Two Cubes Types: a3 þ b3 ¼ (a þ b)(a2  ab þ b2 ) a3  b3 ¼ (a  b)(a2 þ ab þ b2 ) 5.6

x3 þ 8 ¼ x3 þ 23 ¼ (x þ 2)(x2  2x þ 22 ) ¼ (x þ 2)(x2  2x þ 4) a3  27 ¼ a3  33 ¼ (a  3)(a2 þ 3a þ 32 ) ¼ (a  3)(a2 þ 3a þ 9) a6 þ b6 ¼ (a2 )3 þ (b2 )3 ¼ (a2 þ b2 )½(a2 Þ2  a2 b2 þ (b2 )2  ¼ (a2 þ b2 )(a4  a2 b2 þ b4 ) a6  b6 ¼ (a3 þ b3 )(a3  b3 ) ¼ (a þ b)(a2  ab þ b2 )(a  b)(a2 þ ab þ b2 ) a9 þ b9 ¼ (a3 )3 þ (b3 )3 ¼ (a3 þ b3 )½(a3 )2  a3 b3 þ (b3 )2  ¼ (a þ b)(a2  ab þ b2 )(a6  a3 b3 þ b6 ) a12 þ b12 ¼ (a4 )3 þ (b4 )3 ¼ (a4 þ b4 )(a8  a4 b4 þ b8 ) 64x3 þ 125y3 ¼ (4x)3 þ (5y)3 ¼ (4x þ 5y)½(4x)2  (4x)(5y) þ (5y)2  ¼ (4x þ 5y)(16x2  20xy þ 25y2 ) 3 3 (h) (x þ y)  z ¼ (x þ y  z)½(x þ y)2 þ (x þ y)z þ z2  ¼ (x þ y  z)(x2 þ 2xy þ y2 þ xz þ yz þ z2 )

(a) (b) (c) (d ) (e) (f ) (g)

CHAP. 5]

FACTORING

(i) (x  2)3 þ 8y3 ¼ (x  2)3 þ (2y)3 ¼ (x  2 þ 2y)½(x  2)2  (x  2)(2y) þ (2y)2  ¼ (x  2 þ 2y)(x2  4x þ 4  2xy þ 4y þ 4y2 ) 6 3 3 3 (j) x  7x  8 ¼ (x  8)(x þ 1) ¼ (x3  23 )(x3 þ 1) ¼ (x  2)(x2 þ 2x þ 4)(x þ 1)(x2  x þ 1) (k) x8 y  64x2 y7 ¼ x2 y(x6  64y6 ) ¼ x2 y(x3 þ 8y3 )(x3  8y3 ) ¼ x2 y½x3 þ (2y)3 ½x3  (2y)3  ¼ x2 y(x þ 2y)(x2  2xy þ 4y2 )(x  2y)(x2 þ 2xy þ 4y2 ) 6 2 3 2 (l) 54x y  38x y  16y2 ¼ 2y2 (27x6  19x3  8) ¼ 2y2 (27x3 þ 8)(x3  1) ¼ 2y2 ½(3x)3 þ 23 (x3  1) ¼ 2y2 (3x þ 2)(9x2  6x þ 4)(x  1)(x2 þ x þ 1) Grouping of Terms Type: ac þ bc þ ad þ bd ¼ c(a þ b) þ d(a þ b) ¼ (a þ b)(c þ d) 5.7

bx  ab þ x2  ax ¼ b(x  a) þ x(x  a) ¼ (x  a)(b þ x) ¼ (x  a)(x þ b) 3ax  ay  3bx þ by ¼ a(3x  y)  b(3x  y) ¼ (3x  y)(a  b) 6x2  4ax  9bx þ 6ab ¼ 2x(3x  2a)  3b(3x  2a) ¼ (3x  2a)(2x  3b) ax þ ay þ x þ y ¼ a(x þ y) þ (x þ y) ¼ (x þ y)(a þ 1) x2  4y2 þ x þ 2y ¼ (x þ 2y)(x  2y) þ (x þ 2y) ¼ (x þ 2y)(x  2y þ 1) x3 þ x2 y þ xy2 þ y3 ¼ x2 (x þ y) þ y2 (x þ y) ¼ (x þ y)(x2 þ y2 ) x7 þ 27x4  x3  27 ¼ x4 (x3 þ 27)  (x3 þ 27) ¼ (x3 þ 27)(x4  1) ¼ (x3 þ 33 )(x2 þ 1)(x2  1) ¼ (x þ 3)(x2  3x þ 9)(x2 þ 1)(x þ 1)(x  1) 3 3 3 3 (h) x y  y þ 8x  8 ¼ y3 (x3  1) þ 8(x3  1) ¼ (x3  1)(y3 þ 8) ¼ (x  1)(x2 þ x þ 1)(y þ 2)(y2  2y þ 4) 6 6 2 4 4 2 6 2 4 6 4 2 (i) a þ b  a b  a b ¼ a  a b þ b  a b ¼ a2 (a4  b4 )  b2 (a4  b4 ) ¼ (a4  b4 )(a2  b2 ) ¼ (a2 þ b2 )(a2  b2 )(a þ b)(a  b) ¼ (a2 þ b2 )(a þ b)(a  b)(a þ b)(a  b) ¼ (a2 þ b2 )(aþ b)2 (a  b)2 3 2 2 ( j) a þ 3a  5ab þ 2b  b3 ¼ (a3  b3 ) þ (3a2  5ab þ 2b2 ) ¼ (a  b)(a2 þ ab þ b2 ) þ (a  b)(3a  2b) ¼ (a  b)(a2 þ ab þ b2 þ 3a  2b) (a) (b) (c) (d ) (e) (f) (g)

Factors of an + bn 5.8

an þ bn has a þ b as a factor if and only if n is a positive odd integer. Then an þ bn ¼ (a þ b)(an1  an2 b þ an3 b2      abn2 þ bn1 ): a3 þ b3 ¼ (a þ b)(a2  ab þ b2 ) 64 þ y3 ¼ 43 þ y3 ¼ (4 þ y)(42  4y þ y2 ) ¼ (4 þ y)(16  4y þ y2 ) x3 þ 8y6 ¼ x3 þ (2y2 )3 ¼ (x þ 2y2 )½x2  x(2y2 ) þ (2y2 )2  ¼ (x þ 2y2 )(x2  2xy2 þ 4y4 ) a5 þ b5 ¼ (a þ b)(a4  a3 b þ a2 b2  ab3 þ b4 ) 1 þ x5 y5 ¼ 15 þ (xy)5 ¼ (1 þ xy)(1  xy þ x2 y2  x3 y3 þ x4 y4 ) z5 þ 32 ¼ z5 þ 25 ¼ (z þ 2)(z4  2z3 þ 22 z2  23 z þ 24 ) ¼ (z þ 2)(z4  2z3 þ 4z2  8z þ 16) a10 þ x10 ¼ (a2 )5 þ (x2 )5 ¼ (a2 þ x2 )½(a2 )4  (a2 )3 x2 þ (a2 )2 (x2 )2  (a2 )(x2 )3 þ (x2 )4  ¼ (a2 þ x2 )(a8  a6 x2 þ a4 x4  a2 x6 þ x8 ) 7 7 6 5 (h) u þ v ¼ (u þ v)(u  u v þ u4 v2  u3 v3 þ u2 v4  uv5 þ v6 ) (i) x9 þ 1 ¼ (x3 )3 þ 13 ¼ (x3 þ 1)(x6  x3 þ 1) ¼ (x þ 1)(x2  x þ 1)(x6  x3 þ 1)

(a) (b) (c) (d ) (e) (f) (g)

5.9

an  bn has a  b as a factor if n is any positive integer. Then an  bn ¼ (a  b)(an1 þ an2 b þ an3 b2 þ    þ abn2 þ bn1 ): If n is an even positive integer, an  bn also has a þ b as factor. (a) a2  b2 ¼ (a  b)(a þ b)

37

38

FACTORING

(b) (c) (d ) (e) (f ) (g) (h) (i) ( j)

[CHAP. 5

a3  b3 ¼ (a  b)(a2 þ ab þ b2 ) 27x3  y3 ¼ (3x)3  y3 ¼ (3x  y)½(3x)2 þ (3x)y þ y2  ¼ (3x  y)(9x2 þ 3xy þ y2 ) 1  x3 ¼ (1  x)(12 þ 1x þ x2 ) ¼ (1  x)(1 þ x þ x2 ) a5  32 ¼ a5  25 ¼ (a  2)(a4 þ a3  2 þ a2  22 þ a  23 þ 24 ) ¼ (a  2)(a4 þ 2a3 þ 4a2 þ 8a þ 16) y7  z7 ¼ (y  z)(y6 þ y5 z þ y4 z2 þ y3 z3 þ y2 z4 þ yz5 þ z6 ) x6  a6 ¼ (x3 þ a3 )(x3  a3 ) ¼ (x þ a)(x2  ax þ a2 )(x  a)(x2 þ ax þ a2 ) u8  v8 ¼ (u4 þ v4 )(u4  v4 ) ¼ (u4 þ v4 )(u2 þ v2 )(u2  v2 ) ¼ (u4 þ v4 )(u2 þ v2 )(u þ v)(u  v) x9  1 ¼ (x3 )3  1 ¼ (x3  1)(x6 þ x3 þ 1) ¼ (x  1)(x2 þ x þ 1)(x6 þ x3 þ 1) x10  y10 ¼ (x5 þ y5 )(x5  y5 ) ¼ (x þ y)(x4  x3 y þ x2 y2  xy3 þ y4 )(x  y)(x4 þ x3 y þ x2 y2 þ xy3 þ y4 )

Addition and Subtraction of Suitable Terms 5.10

(a) a4 þ a2 b2 þ b4 (adding and subtracting a2 b2 ) ¼ (a4 þ 2a2 b2 þ b4 )  a2 b2 ¼ (a2 þ b2 )2  (ab)2 ¼ (a2 þ b2 þ ab)(a2 þ b2  ab) 4 (b) 36x þ 15x2 þ 4 (adding and subtracting 9x2 ) ¼ (36x4 þ 24x2 þ 4)  9x2 ¼ (6x2 þ 2)2  (3x)2 ¼ ½(6x2 þ 2) þ 3x½(6x2 þ 2)  3x ¼ (6x2 þ 3x þ 2)(6x2  3x þ 2) 4 (c) 64x þ y4 (adding and subtracting 16x2 y2 ) ¼ (64x4 þ 16x2 y2 þ y4 )  16x2 y2 ¼ (8x2 þ y2 )2  (4xy)2 ¼ (8x2 þ y2 þ 4xy)(8x2 þ y2  4xy) 8 (d ) u  14u4 þ 25 (adding and subtracting 4u4 ) ¼ (u8  10u4 þ 25)  4u4 ¼ (u4  5)2  (2u2 )2 ¼ (u4  5 þ 2u2 )(u4  5  2u2 ) ¼ (u4 þ 2u2  5)(u4  2u2  5) Miscellaneous Problems

5.11

(a) x2  4z2 þ 9y2  6xy ¼ (x2  6xy þ 9y2 )  4z2 ¼ (x  3y)2  (2z)2 ¼ (x  3y þ 2z)(x  3y  2z) 2 2 (b) 16a þ 10bc  25c  b2 ¼ 16a2  (b2  10bc þ 25c2 ) ¼ (4a)2  (b  5c)2 ¼ (4a þ b  5c)(4a  b þ 5c) 2 2 (c) x þ 7x þ y  7y  2xy  8 ¼ (x2  2xy þ y2 ) þ 7(x  y)  8 ¼ (x  yÞ2 þ 7(x  y)  8 ¼ (x  y þ 8)(x  y  1) (d ) a2  8ab  2ac þ 16b2 þ 8bc  15c2 ¼ (a2  8ab þ 16b2 )  (2ac  8bc)  15c2 ¼ (a  4b)2  2c(a  4b)  15c2 ¼ (a  4b  5c)(a  4b þ 3c) (e) m4  n4 þ m3  mn3  n3 þ m3 n ¼ (m4  mn3 ) þ (m3 n  n4 ) þ (m3  n3 ) ¼ m(m3  n3 ) þ n(m3  n3 ) þ (m3  n3 ) ¼ (m3  n3 )(m þ n þ 1) ¼ (m  n)(m2 þ mn þ n2 )(m þ n þ 1) Greatest Common Factor and Least Common Multiple

5.12

(a) 9x4 y2 ¼ 32 x4 y2 , 12x3 y3 ¼ 22  3x3 y3 GCF ¼ 3x3 y2 , LCM ¼ 22  32 x4 y3 ¼ 36x4 y3 (b) 48r 3 t4 ¼ 24  3r 3 t4 , 54r 2 t6 ¼ 2  33 r 2 t6 , 60r 4 t2 ¼ 22  3  5r 4 t2 GCF ¼ 2  3r 2 t2 ¼ 6r 2 t2 , LCM ¼ 24  33  5r 4 t6 ¼ 2160r 4 t6 (c) 6x  6y ¼ 2  3(x  y), 4x2  4y2 ¼ 22 (x2  y2 ) ¼ 22 (x þ y)(x  y) GCF ¼ 2(x  y), LCM ¼ 22  3(x þ y)(x  y) (d ) y4  16 ¼ (y2 þ 4)(y þ 2)(y  2), y2  4 ¼ (y þ 2)(y  2), y2  3y þ 2 ¼ (y  1)(y  2) GCF ¼ y  2, LCM ¼ (y2 þ 4)(y þ 2)(y  2)(y  1) (e) 3  52 (x þ 3y)2 (2x  y)4 , 23  32  5(x þ 3y)3 (2x  y)2 , 22  3  5(x þ 3y)4 (2x  y)5 GCF ¼ 3  5(x þ 3y)2 (2x  y)2 , LCM ¼ 23  32  52 (x þ 3y)4 (2x  y)5

CHAP. 5]

39

FACTORING

Supplementary Problems Factor each expression. 5.13

(a) (b) (c) (d) (e) (f ) (g)

3x2 y4 þ 6x3 y3 12s2 t2  6s5 t4 þ 4s4 t 2x2 yz  4xyz2 þ 8xy2 z3 4y2  100 1  a4 64x  x3 8x4  128

(h) 18x3 y  8xy3 (i) (2x þ y)2  (3y  z)2 ( j) 4(x þ 3y)2  9(2x  y)2 (k) x2 þ 4x þ 4 (l ) 4  12y þ 9y2 (m) x2 y2  8xy þ 16 (n) 4x3 y þ 12x2 y2 þ 9xy3

5.14

(a) (b) (c) (d)

m4  4m2  21 a4  20a2 þ 64 4s4 t  4s3 t2  24s2 t3 x2mþ4 þ 5xmþ4  50x4

(e) ( f) (g) (h)

5.15

(a) y3 þ 27 (b) x3  1 (c) x3 y3 þ 8

5.16

(a) xy þ 3y  2x  6 (b) 2pr  ps þ 6qr  3qs

5.17

(a) z5 þ 1

5.18

(a) z4 þ 64 (b) 4x4 þ 3x2 y2 þ y4 (c) x8  12x4 þ 16

5.19

Find the GCF and LCM of each group of polynomials. (a) 16y2 z4 , 24y3 z2 (b) 9r 3 s2 t5 ,12r 2 s4 t3 , 21r 5 s2 (c) x2  3xy þ 2y2 , 4x2  16xy þ 16y2 (d) 6y3 þ 12y2 z, 6y2  24z2 , 4y2  4yz  24z2 (e) x5  x, x5  x2 , x5  x3

2x2 þ 3x þ 1 3y2  11y þ 6 5m3  3m2  2m 6x2 þ 5xy  6y2

(g) y6 þ 1 (h) (x  2)3 þ (y þ 1)3 (i) 8x6 þ 7x3  1

(c) ax2 þ bx  ax  b (d ) x3  xy2  x2 y þ y3 (c) 32  u5

3a4 þ 6a2 b2 þ 3b4 (m2  n2 )2 þ 8(m2  n2 ) þ 16 x2 þ 7x þ 12 y2  4y  5 x2  8xy þ 15y2 2z3 þ 10z2  28z 15 þ 2x  x2

(i) 36z6  13z4 þ z2 ( j ) 12(x  y)2 þ 7(x  y)  12 (k) 4x2nþ2  4xnþ2  3x2

(d ) 8z4  27z7 (e) 8x4 y  64xy4 ( f ) m9  n9

(b) x5 þ 32y5

(o) (p) (q) (r) (s) (t) (u)

(e) z7  2z6 þ z4  2z3 ( f ) m3  mn2 þ m2 n  n3 þ m2  n2

(d) m10  1

(d) m2  4p2 þ 4mn þ 4n2 (e) 6ab þ 4  a2  9b2

(e) 1  z7

( f ) 9x2  x2 y2 þ 4y2 þ 12xy (g) x2 þ y2  4z2 þ 2xy þ 3xz þ 3yz

ANSWERS TO SUPPLEMENTARY PROBLEMS

5.13

(a) (b) (c) (d) (e) (f ) (g)

3x2 y3 (y þ 2x) 2s2 t(6t  3s3 t3 þ 2s2 ) 2xyz(x  2z þ 4yz2 ) 4(y þ 5)(y  5) (1 þ a2 )(1 þ a)(1  a) x(8 þ x)(8  x) 8(x2 þ 4)(x þ 2)(x  2)

(h) (i) ( j) (k) (l) (m) (n)

2xy(3x þ 2y)(3x  2y) (2x þ 4y  z)(2x  2y þ z) (8x þ 3y)(9y  4x) (x þ 2)2 (2  3y)2 (xy  4)2 xy(2x þ 3y)2

(o) ( p) (q) (r) (s) (t) (u)

3(a2 þ b2 )2 (m2  n2 þ 4)2 (x þ 3)(x þ 4) (y  5)(y þ 1) (x  3y)(x  5y) 2z(z þ 7)(z  2) (5  x)(3 þ x)

40

FACTORING

[CHAP. 5

5.14

(a) (b) (c) (d )

(m2  7)(m2 þ 3) (a þ 2)(a  2)(a þ 4)(a  4) 4s2 t(s  3t)(s þ 2t) x4 (xm  5)(xm þ 10)

(e) (f ) (g) (h)

(2x þ 1)(x þ 1) (3y  2)(y  3) m(5m þ 2)(m  1) (2x þ 3y)(3x  2y)

5.15

(a) (b) (c) (d ) (e)

(y þ 3)(y2  3y þ 9) (x  1)(x2 þ x þ 1) (xy þ 2)(x2 y2  2xy þ 4) z4 (2  3z)(4 þ 6z þ 9z2 ) 8xy(x  2y)(x2 þ 2xy þ 4y2 )

(f ) (g) (h) (i)

(m  n)(m2 þ mn þ n2 )(m6 þ m3 n3 þ n6 ) (y2 þ 1)(y4  y2 þ 1) (x þ y  1)(x2  xy þ y2  5x þ 4y þ 7) (2x  1)(4x2 þ 2x þ 1)(x þ 1)(x2  x þ 1)

5.16

(a) (x þ 3)(y  2) (b) (2r  s)(p þ 3q)

5.17

(a) (b) (c) (d ) (e)

(z þ 1)(z4  z3 þ z  z þ 1) (x þ 2y)(x4  2x3 y þ 4x2 y2  8xy3 þ 16y4 ) (2  u)(16 þ 8u þ 4u2 þ 2u3 þ u4 ) (m þ 1)(m4  m3 þ m2  m þ 1)(m  1)(m4 þ m3 þ m2 þ m þ 1) (1  z)(1 þ z þ z2 þ z3 þ z4 þ z5 þ z6 )

5.18

(a) (b) (c) (d )

(z2 þ 4z þ 8)(z2  4z þ 8) (2x2 þ xy þ y2 )(2x2  xy þ y2 ) (x4 þ 2x2  4)(x4  2x2  4) (m þ 2n þ 2p)(m þ 2n  2p)

5.19

(a) (b) (c) (d ) (e)

GCF ¼ 23 y2 z2 ¼ 8y2 z2 , GCF ¼ 3r 2 s2 , GCF ¼ x  2y, GCF ¼ 2(y þ 2z), GCF ¼ x(x  1),

(c) (ax þ b)(x  1) (d ) (x  y)2 (x þ y)

(i) z2 (2z þ 1)(2z  1)(3z þ 1)(3z  1) ( j) (4x  4y  3)(3x  3y þ 4) (k) x2 (2xn þ 1)(2xn  3)

(e) z3 (z  2)(z þ 1)(z2  z þ 1) ( f ) (m þ n)(m  n)(m þ n þ 1)

(e) (2 þ a  3b)(2  a þ 3b) ( f ) (3x þ xy þ 2y)(3x  xy þ 2y) (g) (x þ y þ 4z)(x þ y  z)

LCM ¼ 24  3y3 z4 ¼ 48y3 z4 LCM ¼ 252r 5 s4 t5 LCM ¼ 4(x  y)(x  2y)2 LCM ¼ 12y2 (y þ 2z)(y  2z)(y  3z) LCM ¼ x3 (x þ 1)(x  1)(x2 þ 1)(x2 þ x þ 1)

CHAPTER 6

Fractions 6.1

RATIONAL ALGEBRAIC FRACTIONS

A rational algebraic fraction is an expression which can be written as the quotient of two polynomials, P/Q. P is called the numerator and Q the denominator of the fraction. Thus 3x  4 x2  6x þ 8

and

x3 þ 2y2 x4  3xy þ 2y3

are rational algebraic fractions. Rules for manipulation of algebraic fractions are the same as for fractions in arithmetic. One such fundamental rule is: The value of a fraction is unchanged if its numerator and denominator are both multiplied by the same quantity or both divided by the same quantity, provided only that this quantity is not zero. In such case we call the fractions equivalent. For example, if we multiply the numerator and denominator of (x þ 2)=(x  3) by (x  1) we obtain the equivalent fraction (x þ 2)(x  1) x2 þ x  2 ¼ 2 (x  3)(x  1) x  4x þ 3 provided (x  1) is not zero, i.e., provided x = 1. Similarly, given the fraction (x2 þ 3x þ 2)=(x2 þ 4x þ 3) we may write it as (x þ 2)(x þ 1) (x þ 3)(x þ 1) and divide numerator and denominator by (x þ 1) to obtain (x þ 2)=(x þ 3) provided (x þ 1) is not zero, i.e., provided x = 1. The operation of dividing out common factors of the numerator and denominator is called cancellation and may be indicated by a sloped line thus: (x þ 2)(x þ 1) xþ2 ¼ : (x þ 3)(x þ 1) xþ3 To simplify a given fraction is to convert it into an equivalent form in which numerator and denominator have no common factor (except +1). In such case we say that the fraction is reduced to lowest terms. This 41

42

FRACTIONS

[CHAP. 6

reduction is achieved by factoring numerator and denominator and canceling common factors assuming they are not equal to zero.

Thus

x2  4xy þ 3y2 (x  3y)(x  y) x  3y ¼ ¼ 2 2 (x þ y)(x  y) xþy x y

provided (x  y) = 0:

Three signs are associated with a fraction: the sign of the numerator, of the denominator, and of the entire fraction. Any two of these signs may be changed without changing the value of the fraction. If there is no sign before a fraction, a plus sign is implied. EXAMPLES 6.1. a a a ¼ ¼ , b b b

a a ¼ , b b



a a ¼ b b

Change of sign may often be of use in simplification. Thus x2  3x þ 2 (x  2)(x  1) (x  2)(x  1) x  1 ¼ ¼ ¼ ¼ 1  x: 2x 2x (x  2) 1

6.2

OPERATIONS WITH ALGEBRAIC FRACTIONS

The algebraic sum of fractions having a common denominator is a fraction whose numerator is the algebraic sum of the numerators of the given fractions and whose denominator is the common denominator. EXAMPLES 6.2. 3 4 2 1 3  4  2 þ 1 2 2   þ ¼ ¼ ¼ 5 5 5 5 5 5 5 2 3x þ 4 x2 þ 5 2  (3x þ 4) þ (x2 þ 5) x2  3x þ 3  þ ¼ ¼ x3 x3 x3 x3 x3

To add and subtract fractions having different denominators, write each of the given fractions as equivalent fractions all having a common denominator. The least common denominator (LCD) of a given set of fractions is the LCM of the denominators of the fractions. 7 is the LCM of 4, 5, 10 which is 20, and the LCD of Thus the LCD of 34, 45, and 10 2 3 x , , x2 2x 7

is 14x2 :

EXAMPLES 6.3. 3 4 7 15 16 14 15  16 þ 14 13  þ ¼  þ ¼ ¼ 4 5 10 20 20 20 20 20 2 3 x 2(14)  3(7x)  x(2x2 ) 28  21x  2x3   ¼ ¼ 2 14x2 x 2x 7 14x2 2x þ 1 3 (2x þ 1)(x  1)  3x 2x2  4x  1  ¼ ¼ x(x þ 2) (x þ 2)(x  1) x(x þ 2)(x  1) x(x þ 2)(x  1)

CHAP. 6]

FRACTIONS

43

The product of two or more given fractions produces a fraction whose numerator is the product of the numerators of the given fractions and whose denominator is the product of the denominators of the given fractions. EXAMPLES 6.4. 2 4 15 2  4  15 1   ¼ ¼ 3 5 16 3  5  16 2 x2  9 x  5 (x þ 3)(x  3) x  5  ¼  x2  6x þ 5 x þ 3 (x  5)(x  1) x þ 3 ¼

(x þ 3)(x  3Þ(x  5) x3 ¼ (x  5)(x  1)(x þ 3) x1

The quotient of two given fractions is obtained by inverting the divisor and then multiplying. EXAMPLES 6.5. 3 5 4 8 4

or

3=8 3 4 3 ¼  ¼ 5=4 8 5 10

7 xy 7 xþ2 7 4 ¼  ¼ x2  4 x þ 2 (x þ 2)(x  2) xy xy(x  2)

6.3

COMPLEX FRACTIONS

A complex fraction is one which has one or more fractions in the numerator or denominator, or in both. To simplify a complex fraction:

Method 1 (1) (2)

Reduce the numerator and denominator to simple fractions. Divide the two resulting fractions.

EXAMPLE 6.6. 1 x2  1 2 2 x ¼ x ¼x 1  x ¼x 1¼x1 1 xþ1 x xþ1 xþ1 1þ x x x

Method 2 (1)

Multiply the numerator and denominator of the complex fraction by the LCM of all denominators of the fractions in the complex fraction.

(2)

Reduce the resulting fraction to lowest terms.

EXAMPLE 6.7.   1 1  4 x2 4 1  4x2 (1 þ 2x)(1  2x) x2 x2  ¼ ¼ ¼  1 1 x  2x2 x(1  2x) 2  2 x2 x x ¼

1 þ 2x x

44

FRACTIONS

Solved Problems Reduction of Fractions to Lowest Terms 6.1

(a)

15x2 3  5  x  x 5x ¼ ¼ 12xy 3  4  x  y 4y

(c)

14a3 b3 c2 2a ¼ b 7a2 b4 c2

(b)

4x2 y 22xxy 2x ¼ ¼ 18xy3 2  9  x  y  y2 9y2

(d )

8x  8y 8(x  y) 1 ¼ ¼ (where x  y = 0) 16x  16y 16(x  y) 2

(e)

x3 y  y3 x xy(x2  y2 ) xy(x  y)(x þ y) ¼ ¼xþy ¼ x2 y  xy2 xy(x  y) xy(x  y)

(f)

x2  4xy þ 3y2 (x  3y)(x  y) (x  3y)(x  y) x  3y 3y  x ¼ ¼ ¼ ¼ (y  x)(y þ x) (x  y)(y þ x) yþx yþx y2  x2

(g)

6x2  3xy 3x(2x  y) 3x(2x  y) 3 ¼ ¼ ¼ 2 2 4x y þ 2xy 2xy(y  2x) 2xy(2x  y) 2y

(h)

r 3 s þ 3r 2 s þ 9rs rs(r 2 þ 3r þ 9) rs(r 2 þ 3r þ 9) rs ¼ ¼ ¼ 3 3 3 2 r  27 r 3 (r  3)(r þ 3r þ 9) r  3

(i)

(8xy þ 4y2 )2 (4y½2x þ y)2 16y2 (2x þ y)2 16y(2x þ y) ¼ ¼ 2 ¼ 3 4 3 3 2 2 8x y þ y y(8x þ y ) y(2x þ y)(4x  2xy þ y ) 4x  2xy þ y2

( j)

x2nþ1  x2n y x2n (x  y) x2n (x  y) xn ¼ n ¼ 2 ¼ n 3 nþ3 n 3 3 2 2 x x (x  y ) x (x  y)(x þ xy þ y ) x þ xy þ y2 x y

Multiplication of Fractions 6.2

9 x2  1 9 (x þ 1)(x  1) x  1  ¼  ¼ 3x þ 3 6 3(x þ 1) 6 2

(a)

2x 6y 12xy 4  2¼ 2 2¼ 2 3y x 3x y xy

(c)

x2  4 2xy (x þ 2)(x  2) 2xy 2(x þ 2) ¼  2  ¼ 2 2 2 xy x  4x þ 4 xy y(x  2) (x  2)

(d )

6x  12 y2  1 6(x  2) (y þ 1)(y  1)   ¼ 2 4xy þ 4x 2  3x þ x 4x(y þ 1) (2  x)(1  x)

(b)

6(x  2)(y þ 1)(y  1) 3(y  1) 3(y  1) ¼ ¼ 4x(y þ 1)(x  2)(1  x) 2x(1  x) 2x(x  1)    2 ax þ ab þ cx þ bc x  2ax þ a2 (a þ c)(x þ b) (x  a)(x  a) (e)  ¼ a2  x 2 (a  x)(a þ x) (x þ a)(x þ b) x2 þ (b þ a)x þ ab ¼

¼

(a þ c)(x þ b) (x  a)(x  a) (a þ c)(x  a) (a þ c)(a  x)  ¼ ¼ (x  a)(a þ x) (x þ a)(x þ b) (x þ a)2 (x þ a)2

Division of Fractions 6.3

(a)

5 3 5 11 55 4 ¼  ¼ 4 11 4 3 12

(b)

9 4 9 7 9 4 ¼  ¼ 7 7 7 4 4

(c)

3x 6x2 3x 4 1 4 ¼  2¼ 2 2 6x x 4

[CHAP. 6

CHAP. 6]

45

FRACTIONS

(d )

10xy2 5xy 10xy2 6z3 4 3¼  ¼ 4yz2 6z 3z 3z 5xy

(e)

x þ 2xy 2y þ 1 x þ 2xy 6x x(1 þ 2y) 6x ¼ ¼ ¼2 4   2 2 2 3x 6x 3x 2y þ 1 3x (2y þ 1)

(f)

9  x2 x3  2x2  3x 9  x2 x2 þ 7x þ 6 ¼ 4  x2 þ 7x þ 6 x4 þ 6x3 x4 þ 6x3 x3  2x2  3x ¼

(3  x)(3 þ x) (x þ 1)(x þ 6) 3þx  ¼ 4 3 x (x þ 6) x(x  3)(x þ 1) x

2x2  5x þ 2 (2x2  5x þ 2) 3 (2x  1)(x  2) 3   ¼  ¼  ¼ 3(x  2) 2x  1 1 2x  1 1 2x  1 3  2  x  5x þ 6 x2  5x þ 6 64  x2 (x  3)(x  2) (8  x)(8 þ x) x2 þ 7x  8  (h)    ¼ ¼ x2 þ 7x  8 9  x2 (x þ 8)(x  1) (3  x)(3 þ x) 9  x2 64  x2 (x  2)(8  x) (x  2)(x  8) ¼ ¼ (x  1)(3 þ x) (x  1)(x þ 3)

(g)

Addition and Subtraction of Fractions 6.4

(a)

1 1 2 1 3 1 þ ¼ þ ¼ ¼ 3 6 6 6 6 2

(d)

3t2 4t2 3t2 (3)  4t2 (1) 5t2 t2 ¼  ¼ ¼ 15 5 15 15 3

(b)

5 7 5(4) 7(3) 41 þ ¼ þ ¼ 18 24 72 72 72

(e)

1 1 yþx þ ¼ x y xy

(c)

x 5x x(7) þ 5x(2) 17x þ ¼ ¼ 6 21 42 42

(f)

3 4 3(3y) þ 4(x) 9y þ 4x þ ¼ ¼ x 3y 3xy 3xy

(g)

5 3 5(2x)  3(1) 10x  3  2¼ ¼ 2x 4x 4x2 4x2

(h)

3a 2b 3a(a) þ 2b(b) 3a2 þ 2b2 þ ¼ ¼ bc ac abc abc

(i)

3t  1 5  2t (3t  1)3 þ (5  2t)2 9t  3 þ 10  4t 5t þ 7 þ ¼ ¼ ¼ 10 15 30 30 30

3 2 2 3x(x þ 1)  2x2 þ 2(x þ 1) x2 þ 5x þ 2  þ 2¼ ¼ 2 x xþ1 x x2 (x þ 1) x (x þ 1) 5 10 5(x2  9)  5(x  3) þ 10 5(x2  x  4) þ ¼ ¼ (k) 5  x þ 3 x2  9 x2  9 x2  9

( j)

(l)

3 2 y 3(y þ 2)  2(y  2)  y 10   ¼ ¼ 2 y  2 y þ 2 y2  4 y2  4 y 4

Complex Fractions 6.5

(a)

5=7 5 4 20 ¼  ¼ 3=4 7 3 21

(b)

2=3 2 1 2 ¼  ¼ 7 3 7 21

(c)

10 10 6 60 ¼  ¼ ¼ 12 5=6 1 5 5

46

FRACTIONS

   2 5 4 5 9 þ þ 3 6 6 6 6 ¼98¼4 ¼ ¼  (d ) 3 3 3 6 3 8 8 8   xþy xþy x xþy 3x2 ¼ ¼ (e)   2 xy 3x x  y 3x(x  y) x     1 1 þ1 þ1 x 1þx x x ¼  ¼ (h)  1x 1 1 1 1 x x x

[CHAP. 6



 (f )

 2 2 1 2 ab ¼  ¼ ab a  b a  b (a  b)2

2a 2a x þ 1 ¼  ¼ 2(x þ 1) a 1 a xþ1

(g) 

Supplementary Problems Show that: 6.6

6.7

6.8

6.9

6.10

(a)

24x3 y2 4x2 ¼ 18xy3 3y

(d )

4x2  16 4(x þ 2) ¼ x2  2x x

(g)

ax4  a2 x3  6a3 x2 x(x þ 2a) ¼ 9a4 x  a2 x3 a(x þ 3a)

(b)

36xy4 z2 12yx ¼ 15x4 y3 z 5x3

(e)

y2  5y þ 6 3  y ¼ 4  y2 yþ2

(h)

xy  y2 1 ¼ x4 y  xy4 x(x2 þ xy þ y2 )

(c)

5a2  10ab ¼ 5a a  2b

(f)

(x2 þ 4x)2 x2 (x þ 4) ¼ x2 þ 6x þ 8 xþ2

(i)

3a2 2b4 b  ¼ 4b3 9a3 6a

(a)

8xyz2 9xy2 z 6y ¼ 2 3  3x3 y2 z 4xz5 x z

(d )

x2  4y2 2y2  2 2(x þ 2y)(y  1) ¼  2 3xy þ 3x 2y þ xy  x2 3x(x þ y)

(b)

xy2 x2  y2 x þ y  3 2 ¼ 2x2 2x  2y x y

(e)

y2  y  6 y2 þ 3y  4 (y þ 2)(y þ 4)  ¼ y2  2y þ 1 9y  y3 y(y  1)(y þ 3)

(c)

x2 þ 3x 2x2 þ 2x x2  4x þ 3 1 ¼   4x2  4 x2  9 x2 2

(f)

t3 þ 3t2 þ t þ 3 8  t3 (t þ 3)(t2 þ 2t þ 4)  ¼ 4t2  16t þ 16 t3 þ t 4t(2  t)

24x3 y2 8x2 y3 9xz2 x2  4y2 x2  xy  6y2 y(x  2y) 4 ¼ (c) 2 4 ¼ 2 4 5z 15z y x þ xy y2 þ xy x(x  3y)  2   2  y  3y þ 2 x y þ xy2 6x2  x  2 (y  1)(y þ 3) xy xy y2 þ 4y  21  ¼ (2x þ 1)2 (c) (b)   ¼ ¼ (a)  3x  2 (y  2)(y þ 7) xy xþy 4  4y þ y2 2x þ 1 9  y2 (a)

3x 9x 2 4 ¼ 8y 16y 3

(a)

2x x x  ¼ 3 2 6

(e)

(b)

4 5 1  ¼ 3x 4x 12x

(f)

(c)

3 8 3  16y  ¼ 2y2 y 2y2

(g)

(d )

x þ y2 x  1 y2 þ 1¼ 2 2 x x x

(h)

(b)

1 1 x x þ  ¼ x þ 2 x  2 x2  4 x2  4

r2

r1 rþ2 1 r 2 þ 4r þ 12  2 þ ¼ þ r  6 r þ 4r þ 3 3r  6 3(r þ 3)(r  2)(r þ 1)

2x2

x xy y 3x2 þ xy  2 þ 2 ¼ 2 2 2 þ 3xy þ y y  4x 2x þ xy  y (2x þ y)(2x  y)(x þ y)

a b c þ þ ¼0 (c  a)(a  b) (a  b)(b  c) (b  c)(c  a)

CHAP. 6]

6.11

(a)

FRACTIONS

xþy ¼ xy 1 1 þ x y

1 1 (b) 2 x ¼ 2 2x þ x x

  xþ1 x1  x1 xþ1 ¼2 (d )  1 1 þ xþ1 x1

2þ



 2y yþ y2  ¼yþ2 (c)  4 1þ 2 y 4

(e)

x 0

1¼xþy 1 1@ xA 1þ y

(f ) 2 

0

2 2

1 ¼ 2x2

1@ 2A 2 2 x

47

CHAPTER 7

Exponents 7.1

POSITIVE INTEGRAL EXPONENT

If n is a positive integer, an represents the product of n factors each of which is a. Thus a4 ¼ a  a  a  a: In an , a is called the base and n the exponent or index. We may read an as the “nth power of a” or “a to the nth.” If n ¼ 2 we read a2 as “a squared”; a3 is read “a cubed.” EXAMPLES 7.1.

7.2

x3 ¼ x  x  x,

25 ¼ 2  2  2  2  2 ¼ 32,

(3)3 ¼ (3)(3)(3) ¼ 27

NEGATIVE INTEGRAL EXPONENT

If n is a positive integer, we define an ¼

1 an

assuming a = 0:

EXAMPLES 7.2. 24 ¼

7.3

1 1 ¼ , 24 16

1 ¼ 33 ¼ 27, 33

4x2 ¼

4 , x2

(a þ b)1 ¼

1 (a þ b)

ROOTS

If n is a positive integer and if a and b are such that an ¼ b, then a is said to be an nth root of b. pffiffiffi If b is positive, there is only one positive number a such that an ¼ b. We write this positive number n b and call it the principal nth root of b. EXAMPLE 7.3. p ffiffiffi 4 16 ¼ þ2.

ffiffiffi p 4 16 is that positive number which when raised to the 4th power yields 16. Clearly this is þ2, so we write

EXAMPLE 7.4. The number 2 when raised to the 4th power also yields 16. We call 2 a 4th root of 16 but not the principal 4th root of 16. If b is negative, there is no positive nth root ofp b,ffiffiffibut there is a negative nth root of b if n is odd. We call this negative number the principal nth root of b and we write it n b. p ffiffiffiffiffiffiffiffiffi 3 EXAMPLE 27 is that number which raised to the third power (or cubed) yields 27. Clearly this is 3 and so we ffiffiffiffiffiffiffiffiffi 7.5. p 3 write 27 ¼ 3 as the principal cube root of 27.

48

CHAP. 7]

49

EXPONENTS

EXAMPLE 7.6. If n is even, as in

p ffiffiffiffiffiffiffiffiffi 4 16, there is no principal nth root in terms of real numbers.

Note. In advanced mathematics it can be shown that there are exactly n values of a such that an ¼ b, b = 0, provided we allow imaginary (or complex) numbers. 7.4

RATIONAL EXPONENTS

If m and n are positive integers we define pffiffiffiffiffiffi am=n ¼ n am (assume a  0 if n is even) EXAMPLES 7.7. 43=2 ¼

pffiffiffiffiffi pffiffiffiffiffi 43 ¼ 64 ¼ 8,

(27)2=3 ¼

p ffiffiffiffiffiffiffiffiffiffi 3 (27)2 ¼ 9

If m and n are positive integers we define am=n ¼

1 am=n

EXAMPLES 7.8. 82=3 ¼

1 1 1 1 ffiffiffiffiffi ¼ p ffiffiffiffiffi ¼ , ¼p 3 3 82=3 64 4 82

x5=2 ¼

1 1 ¼ pffiffiffiffiffi x5=2 x5

We define a0 ¼ 1 if a = 0. EXAMPLES 7.9. 100 ¼ 1,

7.5

(3)0 ¼ 1,

(ax)0 ¼ 1

(if ax = 0)

GENERAL LAWS OF EXPONENTS

If p and q are real numbers, the following laws hold. A.

ap  aq ¼ apþq

EXAMPLES 7.10. 5p3ffiffiffi  57 ¼ 53þ7 ¼ 54 ¼ 625, 21=2  25=2 ¼ 23 ¼ 8 23  22 ¼ 23þ2 ¼ 25 ¼ 32, 39  32  33 ¼ 34 ¼ 81 31=3  31=6 ¼ 31=3þ1=6 ¼ 31=2 ¼ 3,

B.

(ap )q ¼ apq

EXAMPLES 7.11. (24 )3 ¼ 2(4)(3) ¼ 212 , (x5 )4 x(5)(4) ¼ x20 ,

C.

ap ¼ apq aq

(51=3 )3 ¼ 5(1=3)(3) ¼ 51 ¼ 1=5, (a2=3 )3=4 ¼ a(2=3)(3=4) ¼ a1=2

(32 )0 ¼ 3(2)(0) ¼ 30 ¼ 1

a=0

EXAMPLES 7.12. 26 32 x1=2 ¼ 264 ¼ 22 ¼ 4, ¼ 324 ¼ 36 , ¼ x1=2(1) ¼ x3=2 24 34 x1 pffiffiffiffiffiffiffiffiffiffiffiffiffi (x þ 15)4=3 ¼ (x þ 15)4=35=6 ¼ (x þ 15)1=2 ¼ x þ 15 5=6 (x þ 15)

D.

(ab)p ¼ ap bp

50

EXPONENTS

[CHAP. 7

EXAMPLES 7.13. (2  3)4 ¼ 24  34 ,

E.

 p a ap ¼ p b b

EXAMPLES 7.14.

7.6

(2x)3 ¼ 23 x3 ¼ 8x3 ,

(3a)2 ¼ 32 a2 ¼

pffiffiffi (4x)1=2 ¼ 41=2 x1=2 ¼ 2x1=2 ¼ 2 x

1 9a2

b=0

 5  2 3 2 25 32 x (x2 )3 x6 y9 , ¼ 5¼ ¼ ¼ ¼ 3 y3 3 243 (y3 )3 y9 x6  3 1=3 3 1=3 5 51 22 4 ¼ (56 )1=3 ¼ 2 ¼ 1 ¼ 5 26 2 5 (2 )

SCIENTIFIC NOTATION

Very large and very small numbers are often written in scientific notation when they are used in computation. A number is written in scientific notation by expressing it as a number N times a power of 10 where 1  N , 10 and N contains all of the significant digits on the number. EXAMPLES 7.15.

Write each number in scientific notation. (a) 5 834 000, (b) 0.028 031, (c) 45.6.

(a) 5 834 000 ¼ 5:834  106 (b) 0:028 031 ¼ 2:8031  102 (c) 45:6 ¼ 4:56  101

We can enter the number 3:1416  103 into a calculator by using the EE key or the EXP key. When we enter 3.1416, press the EE key, and then enter 3 followed by pressing the ENTER key or the ¼ key, we get a display of 3141.6. Similarly, we can enter 4:902  102 by entering 4.902, pressing the EE key, and then entering 22 followed by pressing the ENTER key to obtain a display of 0.049 02. The exponent can usually be any integer from 99 to 99. Depending on the number of digits in the number and the exponent used, a calculator may round off the number and/or leave the result in scientific notation. How many digits you can have in the number N varies from calculator to calculator, as does whether or not the calculator displays a particular result in scientific notation or in standard notation. Calculators sometimes display the result in scientific notation, such as 3.69E-7 or 3:6907 . In each case the answer is to be interpreted as 3:69  107 . Calculators display the significant digits in the result followed by the power of 10 to be used. When entering a number in scientific notation into your calculator as part of a computation, press the operation sign after each number until you are ready to compute the result. EXAMPLE 7.16. Compute (1:892  108 )  (5:34  103 ) using a calculator. Enter 1.892, press the EE key, enter 8, press the  sign, enter 5.34, press the EE key, enter 3, and press the ENTER key and we get a display of 1 010 328. (1:892  108 )  (5:34  103 ) ¼ 1 010 328

Solved Problems Positive Integral Exponent 7.1

(a) 23 ¼ 2  2  2 ¼ 8 (b) (3)4 ¼ (3)(3)(3)(3) ¼ 81  5       2 2 2 2 2 2 32 (c) ¼ ¼ 3 3 3 3 3 3 243

(d ) (3y)2 (2y)3 ¼ (3y)(3y)(2y)(2y)(2y) ¼ 72y5 (e) (3xy2 )3 ¼ (3xy2 )(3xy2 )(3xy2 ) ¼ 27x3 y6

CHAP. 7]

51

EXPONENTS

Negative Integral Exponent 7.2

(a) 23 ¼

1 1 ¼ 23 8

(b) 31 ¼

1 1 ¼ 1 3 3

4 ¼ 4x2 y2 x2 y2  3  3 3 1 4 64 (i) ¼ ¼ ¼ 3 4 3 27 (3=4)  3  3 x 1 y y3 ( j) ¼ ¼ ¼ 3 y x x (x=y)3  1 2 100 ¼ 50 (k) (0:02)1 ¼ ¼ 100 2 (h)

  1 1 ¼ 2 4 4   1 2 ¼ 2 2 ¼  2 b b

(c)  4(4)2 ¼  4 (d) 2b2

(e) ð2bÞ2 ¼

(l)

ab4 a  a2 a3 ¼ ¼ a2 b b  b4 b5

( f ) 5  103

(m)

(x  1)2 (x þ 3)1 (2x  4)(x þ 5)3 ¼ (2x  4)1 (x þ 5)3 (x  1)2 (x þ 3)

(g)

7.3

1 1 ¼ (2b)2 4b2   1 5 1 ¼ ¼ ¼5 3 10 1000 200

8 ¼ 8  102 ¼ 800 102

Rational Exponents pffiffiffiffiffi pffiffiffiffiffi (a) (8)2=3 ¼ 3 82 ¼ 3 64 ¼ 4 (c) (x3 )1=3 ¼

7.4

7.5

p ffiffiffiffiffiffiffiffi 3 x3 ¼ x

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi (b) (8)2=3 ¼ 3 (8)2 ¼ 3 64 ¼ 4  1=2 rffiffiffiffiffi  2=3 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi rffiffiffiffiffi 1 1 1 1 1 1 3 3 1 (e)  ¼ ¼ ¼ ¼ ¼  (d) 16 16 4 8 8 64 4

1 1 1 1 1 1 ffiffiffi ffiffiffiffiffiffiffiffi ¼ ¼ ¼p ¼p (d ) (x3 )1=3 ¼ 3 3 x1=3 x x (x3 )1=3 x3 x 1 1 1 1 1 2=3 2=3 ffiffiffiffiffi ¼ ffiffiffiffiffiffiffiffiffiffiffi ¼ 1 (e) (1) (b) (8) ¼ 2=3 ¼ p ¼ ¼p 3 3 8 (1)2=3 82 4 (1)2 1 1 1 1 2=3 2=3 ffiffiffiffiffiffiffiffiffiffiffiffi ¼ ( f ) (1) (c) (8) ¼ ¼p ¼  2=3 ¼ 1 3 1 (8)2=3 ð8Þ2 4 1 1 1 1 3=5 ffiffiffiffiffiffiffi ¼  ffiffiffiffiffiffiffiffiffiffiffi ¼  p ¼1 ¼ ¼ p (g) (1) 5 5 1 (1)3=5 1 (1)3 (a) x1=3 ¼

70 ¼ 1, (3)0 ¼ 1, (2=3)0 ¼ 1 (e) 4  100 ¼ 4  1 ¼ 4 0 (x  y) ¼ 1, if xy=0 ( f ) (4  10)0 ¼ (40)0 ¼ 1 0 x=0 (g) (1)0 ¼ 1 3x ¼ 3  1 ¼ 3, if 0 3x = 0, i.e. if x = 0 (h) (1)0 ¼ 1 (3x) ¼ 1, if 0 0 if 3x = 0 and 4y = 0, i.e. if (3x) (4y) ¼ 1  1 ¼ 1, 2(3x þ 2y  4)0 ¼ 2(1) ¼ 2, if 3x þ 2y  4 = 0 (5x þ 3y) 5x þ 3y ¼ 5x þ 3y, if 5x þ 3y = 0 ¼ (k) 0 1 (5x þ 3y) (l ) 4(x2 þ y2 )(x2 þ y2 )0 ¼ 4(x2 þ y2 )(1) ¼ 4(x2 þ y2 ), if x2 þ y2 = 0

(a) (b) (c) (d) (i) ( j)

x = 0,

y=0

General Laws of Exponents 7.6

(a) (b) (c) (d) (e) (f)

ap  aq ¼ apþq a3  a5 ¼ a3þ5 ¼ a8 34  35 ¼ 39 anþ1  an2 ¼ a2n1 x1=2  x1=3 ¼ x1=2þ1=3 ¼ x5=6 x1=2  x1=3 ¼ x1=21=3 ¼ x1=6

(g) (h) (i) ( j) (k) (l)

107  103 ¼ 1073 ¼ 104 (4  106 )(2  104 ) ¼ 8  1046 ¼ 8  102 ax  ay  az ¼ axþyz pffiffiffiffiffiffiffiffiffiffiffi ( x þ y)(x þ y) ¼ (x þ y)1=2 (x þ y)1 ¼ (x þ y)3=2 101:7  102:6 ¼ 104:3 104:1  103:5  10:1 ¼ 104:1þ3:5:1 ¼ 10:7

52

EXPONENTS

[CHAP. 7

 3=2  2=3  3=22=3  5=6 b b b b (m)  ¼ ¼ a a a a 1=2 rffiffiffiffiffiffiffiffiffiffiffi  1  1=2  1=2  x x x xþy xþy (n) ¼ ¼ ¼ xþy xþy xþy x x (o) (x2 þ 1)5=2 (x2 þ 1)0 (x2 þ 1)2 ¼ (x2 þ 1)5=2þ0þ2 ¼ (x2 þ 1)1=2 ¼

7.7

7.8

1 1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi (x2 þ 1)1=2 x2 þ 1

( f ) (49)3=2 ¼ (72 )3=2 ¼ 723=2 ¼ 73 ¼ 343 (ap )q ¼ apq (x3 )4 ¼ x34 ¼ x12 (g) (31=2 )2 ¼ 31 ¼ 3 mþ2 n (mþ2)n mnþ2n (a ) ¼ a ¼a (h) (u2 )3 ¼ u(2)(3) ¼ u6 (103 )2 ¼ 1032 ¼ 106 (i) (81)3=4 ¼ (34 )3=4 ¼ 33 ¼ 27 pffiffiffiffiffiffiffiffiffiffiffi 3 2 32 6 (10 ) ¼ 10 ¼ 10 ( j) ( x þ y)5 ¼ ½(x þ y)1=2 5 ¼ (x þ y)5=2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( 3 x3 þ y3 )6 ¼ ½(x3 þ y3 )1=3 6 ¼ (x3 þ y3 )1=36 ¼ (x3 þ y3 )2 ffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffi pffiffiffiffiffi p pffiffiffi 6 3 a2 ¼ 6 a2=3 ¼ (a2=3 )1=6 ¼ a1=9 ¼ 9 a (l)

(a) (b) (c) (d) (e) (k)

ap y2=3 pq ¼ a (g) ¼ y2=31=3 ¼ y1=3 aq y1=3 a5 z1=2 (b) 3 ¼ a53 ¼ a2 (h) 3=4 ¼ z1=23=4 ¼ z1=4 a z 74 (x þ y)3aþ1 43 1 (c) 3 ¼ 7 ¼7 ¼7 (i) ¼ (x þ y)a4 7 (x þ y)2aþ5 p2nþ3 8  102 8 (d ) nþ1 ¼ p(2nþ3)(nþ1) ¼ p nþ2 ( j) ¼  102þ6 ¼ 4  108 6 2 p 2  10 2 102 9  10 9 (e) ¼ 1025 ¼ 103 (k) ¼  1024 ¼ 3  106 3 105 3  104 mþ3 3 1=2 x ab ( f ) m1 ¼ x4 (l) ¼ a2 b1 ¼ a 2 b x ab3=2   4x3 y2 z3=2 4 3þ1=2 2þ4 3=21 ¼ y z ¼ 2x7=2 y2 z5=2 (m) x 1=2 4 2x y z 2 ffiffiffiffiffipffiffiffipffiffiffiffiffiffiffi p 3 8 x2 4 y 1=z 8x2=3 y1=4 z1=2 (n) ¼  4x1=3 y9=4 z1 pffiffiffipffiffiffiffiffipffiffi ¼ 2 3 x y5 z 2x1=3 y5=2 z1=2 (a)

7.9

(a) (b) (e) (f) (g)

7.10

(a)

(ab)p ¼ ap bb (c) (3  102 )4 ¼ 34  108 ¼ 81  108 (2a)4 ¼ 24 a4 ¼ 16a4 (d ) (4x8 y4 )1=2 ¼ 41=2 (x8 )1=2 (y4 )1=2 ¼ 2x4 y2 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1=3 12 1=3 6 1=3 12 6 1=3 12 6 64a b ¼ (64a b ) ¼ (64) (a ) (b ) ¼ 4a4 b2 (x2n y1=2 zn1 )2 ¼ x4n y1 z2n2 (27x3p y6q z12r )1=3 ¼ (27)1=3 (x3p )1=3 (y6q )1=3 (z12r )1=3 ¼ 3xp y2q z4r

 p a ap ¼ p b b  4 2 24 16 (b) ¼ 4¼ 3 81 3  3 3 3a (3a) 27a3 (c) ¼ ¼ 4b (4b)3 64b3

 2 n x x2n ¼ 3n 3 y y  mþ1 m 2 a am þm (e) ¼ m b b  2 3=2 2 3=2 a (a ) a3 (f) ¼ ¼ b4 (b4 )3=2 b6 (d )

(where a  0, b = 0)

CHAP. 7]

53

EXPONENTS

 3  3 2 5 125 ¼ ¼ 5 2 8  3 1=3  6 1=3 5 2 22 4 ¼ ¼ ¼ 5 26 53 5 sffiffiffiffiffiffiffiffiffi   1=3 3n 8x3n (8x3n )1=3 81=3 xn 2xn 3 8x ¼ ¼ 1=3 2 ¼ 2 ¼ 1=3 6 6 6 27y 27y 27 y 3y (27y )  1=3 3=2 a (a1=3 )3=2 a1=2 ¼ 1=3 3=2 ¼ 1=2 1=3 x x (x )  1=3 2 3=2 x y (x1=3 y2 )3=2 (x1=3 )3=2 (y2 )3=2 x1=2 y3 ¼ ¼ ¼ 6 z4 z6 z (z4 )3=2 ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffi ffi  1=5 3=4 1=2 p ffiffiffi 4 3 5 x1=5 y3=4 x y x1=10 y3=8 x y ffiffiffiffi p ¼ ¼ ¼ 3 2 2=3 2=3 x z z1=3 z

(g) (h)

(i)

( j)

(k) (l)

Miscellaneous Examples 7.11

1 1 1 7 (a) 23 þ 22 þ 21 þ 20 þ 21 þ 22 þ 23 ¼ 8 þ 4 þ 2 þ 1 þ þ þ ¼ 15 2 4 8 8 1 1 5 (b) 43=2 þ 41=2 þ 41=2 þ 43=2 ¼ 8 þ 2 þ þ ¼ 10 2 8 8 4x0 (c) 4 ¼ 4(1)(24 ) ¼ 4  16 ¼ 64 2 (d) 104 þ 103 þ 102 þ 101 þ 100 þ 101 þ 102 ¼ 10 000 þ 1000 þ 100 þ 10 þ 1 þ 0:1 þ 0:01 ¼ 11 111:11 (e) 3  103 þ 5  102 þ 2  101 þ 4  100 ¼ 3524 43n (22 )3n 26n ¼ n ¼ n ¼ 26nn ¼ 25n 2n 2 2 p ffiffiffiffiffiffiffiffiffiffiffi 3 0:125 0:5 1=3 ¼1 (g) (0:125) (0:25)1=2 ¼ pffiffiffiffiffiffiffiffiffi ¼ 0:5 0:25

(f )

7.12

(a) Evaluate 4x2=3 þ 3x1=3 þ 2x0 when x ¼ 8: 4  82=3 þ 3  81=3 þ 2  80 ¼

4 4 þ 3  81=3 þ 2  80 ¼ þ 3  2 þ 2  1 ¼ 9 82=3 4

(b) Evaluate (3)2 (2x)3 (x þ 1)2 when x ¼ 2. 

(3)2 (4)3 32

7.13

(a)

1 9 4 ¼ 1 32

3 

1 ¼9  4

20  22 1  1=22 1  1=4 3=4 1 ¼ ¼ ¼ ¼ 2 2  2=22 2  2=4 6=4 2 2  2(2)

3 (9) ¼ 

81 64

54

EXPONENTS

[CHAP. 7



(b)

(c)

(d) (e)

   2 2þa þ1 2a1 þ a0 2þa 2 a  a   a ¼ (2 þ a)a ¼ 2a þ a2 ¼ ¼   ¼ 2 1 1 a a a2 a2  0 1  1  0 1  1=3 1 2 1 1 2 8 2 ¼ 2 or ¼ ¼ ¼ ¼ ¼2 1=3 1=3 0 2 1=2 1 8 8 2  2  2 1 1 1 80 (3)2 ¼ (3)2   ¼9 ¼ 3 3 9 9 "  #2=5       1 2=3 1 2=5 1 2=5 1 5 2=3 3 2=3  þ  ¼ (27) þ  5 ¼ ½(3)  þ  27 32 2 2  2 1 37 ¼ (3) þ  ¼ 2 4 2

7.14

(a)

(3a)3  3a2=3 (3a)3  3  (2a)2 27a3  3  4a2 324a5 ¼ 324a4 ¼ ¼ ¼ a2=3  a1=3 a2=3þ1=3 a (2a)2  a1=3

3 (x2 )3  (x1=3 )9 x6  x3 x63 ¼ x ¼ x12 ¼ ¼ 3=215=2 3 5 3=2 15=2 9 x x x x (x1=2 )  (x3=2 )     m  n  m  n  1 1 xy þ 1 xy  1 ð xy þ 1Þm ð xy  1Þn  xþ  x  y y y y ym yn (c)     m  n ¼  m  n ¼  1 1 xy þ 1 xy  1 ð xy þ 1Þm ð xy  1Þn  yþ  y  x x x x xm xn ! ð xy þ 1Þm ð xy  1Þn  mþn ymþn xmþn x ¼ ¼ ¼  mþn m n y y ð xy þ 1Þ ð xy  1Þ

(b)

xmþn

7.15

(d )

3pqþq 32p 3pqþqþ2p  ¼ ¼ 3(pqþqþ2p)(pqþpþ2q) ¼ 3pq 3pqþp 32q 3pqþpþ2q

(a)

(x3=4  x1=2 )1=3 (x5=4 )1=3 x5=12 ¼ 2 1=2 ¼ y (y2=3  y4=3 )1=2 (y )

(b)

(x3=4 )2=3  (y5=4 )2=5 x1=2  y1=2 (x1=4 )2  (y1=4 )2 (x1=4 þ y1=4 )(x1=4  y1=4 ) ¼ ¼ ¼ ¼ x1=4  y1=4 x1=4 þ y1=4 x1=4 þ y1=4 (x3=4 )1=3 þ (y2=3 )3=8 x1=4 þ y1=4

(c)

1 1 þ ¼ pq 1þx 1 þ xqp

(d )

x3n  y3n (xn )3  (yn )3 (xn  yn )(x2n þ xn yn þ y2n ) ¼ ¼ ¼ x2n þ xn yn þ y2n xn  yn xn  yn xn  yn

(e)

1 1 xq xp xq þ xp  p þ  q ¼ q þ ¼ ¼1 x x x þ xp xp þ xq xq þ xp 1þ q 1þ p x x

p ffiffiffiffiffiffiffiffiffiffi ffi a aa2 a ¼ ½aa(a1) 1=a ¼ aa1 32

9

( f ) 22 ¼ 22 ¼ 2512

CHAP. 7]

7.16

55

EXPONENTS

(a) (0:004)(30 000)2 ¼ (4  103 )(3  104 )2 ¼ (4  103 )(32  108 ) ¼ 4  32  103þ8 ¼ 36  105 or 3 600 000 (b)

48 000 000 48  106 ¼ ¼ 4  10 62 ¼ 4  104 1200 12  102

(c)

0:078 78  103 ¼ ¼ 6:5  103þ5 ¼ 6:5  102 0:00012 12  105

or

40 000 or

650

(80 000 000)2 (0:000 003) (8  107 )2 (3  106 ) 82  3 1014  106 ¼ ¼  ¼ 2  1019 (600 000)(0:0002)4 (6  105 )(2  104 )4 6  24 105  1016 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3 4 12 ) (36  104 )  104 3 256(36) 10 3 (0:004) (0:0036) 3 (4  10  (e) ¼ ¼ 2 2 4 108 144 (120 000) (12  10 ) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ¼ 3 64  1024 ¼ 4  108 (d)

7.17

For what real values of the variables involved will each of the following operations be valid and yield real numbers? pffiffiffiffiffi (a) x2 ¼ (x2 )1=2 ¼ x1 ¼ x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (b) a2 þ 2a þ 1 ¼ (a þ 1)2 ¼ a þ 1 a2  b2 (a1 )2  (b1 )2 (a1 þ b1 )(a1  b1 ) ¼ a1 þ b1 ¼ ¼ (a1  b1 ) a1  b1 a1  b1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d) x4 þ 2x2 þ 1 ¼ (x2 þ 1)2 ¼ x2 þ 1 (c)

pffiffiffiffiffiffiffiffiffiffiffi x1 (x  1)1 (e) pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ (x  1)11=2 ¼ (x  1)1=2 ¼ x  1 1=2 x  1 (x  1) SOLUTION

pffiffiffiffiffi pffiffiffiffiffi (a) When x is a realpnumber, x2 must be positive or zero. Assuming x2 ¼ x were ffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi for all x, then if x ¼ 1 ptrue p ffiffi ffi or 1 ¼ 1, i.e. 1 ¼ 1, a contradiction. Thus x2 ¼ x cannot be true for all we would have (1)2 ¼p1 pffiffiffiffiffi ffiffiffiffiffi 2 2 values of x.pWe ffiffiffiffiffi will have x ¼ x if x  0. If x  0 we have x ¼ x. A result valid both for x  0 and x  0 may be x2 ¼ jxj (absolute value of x). pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi zero and thus (b) a2 þ 2a þ 1 must be positive or ffi will equal a þ 1 if a þ 1  0, i.e. if a  1. A result valid for pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi all real values of a is given by a2 þ 2a þ 1 ¼ ja þ 1j. (c) (a2  b2 )=(a1  b1 ) is not defined if a or b or both is equal to zero. Similarly it is not defined if the denominator a1  b1 ¼ 0, i.e. if a1 ¼ b1 or a ¼ b. Hence the result (a2  b2 )=(a1  b1 ) ¼ a1 þ b1 is valid if and only if a = 0, b = 0, and a = b. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d ) x4 þ 2x2 þ 1 must be positive or zero and will equal x2 þ 1 if x2 þ 1  0. Since x2 þ 1 is greater than zero for all real numbers x, the result is valid for all real values of x. pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (e) x  1 will not be a real number if x  1 , 0, i.e.p ifffiffiffiffiffiffiffiffiffiffi x , ffi1. Also, (x  pffiffiffiffiffiffiffiffiffiffi ffi 1)=( x  1) will not be defined if the denominator is zero, i.e. if x ¼ 1. Hence (x  1)=( x  1) ¼ x  1 if and only if x . 1.

7.18

A student was asked to evaluate the expression x þ 2y þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (x  2y)2 for x ¼ 2, y ¼ 4. She wrote

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x þ 2y þ (x  2y)2 ¼ x þ 2y þ x  2y ¼ 2x and thus obtained the value 2x ¼ 2(2) ¼ 4 for her answer. Was she correct?

56

EXPONENTS

[CHAP. 7

SOLUTION Putting x ¼ 2, y ¼ 4 in the given expression, we obtain qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi x þ 2y þ (x  2y)2 ¼ 2 þ 2(4) þ (2  8)2 ¼ 2 þ 8 þ 36 ¼ 2 þ 8 þ 6 ¼ 16: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The student made thepmistake of writing (x  2y)2 ¼ x  2y which is true only if x  2y. If x  2y, (x  2y)2 ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2y  x. In all cases, (x  2y) ¼ jx  2yj. The required simplification should have been x þ 2y þ 2y  x ¼ 4y, which does give 16 when y ¼ 4.

Supplementary Problems Evaluate each of the following. 7.19

7.20

7.21

(a) 34

(e) (4x)2

(i)

82=3 (8)2=3 81=3

(b) (2x)3 ( f ) (2y1 )1  3 3y 31 x2 y4 (c) (g) 2 3 3 2 x y 4

( j) (a3 b3 )2=3

(n) xy  x4y

(k) 3(1)1=5 (4)1=2

(o) 3y2=3  y4=3

(d ) 43

(l) (103 )0

( p) (4  103 )(3  105 )(6  104 )

(h) (16)1=4

(x þ y)2=3 (x þ y)1=6 ½(x þ y)2 1=4

(a)

23  22  24 21  20  23

(d )

(b)

10xþy  10yx  10yþ1 10yþ1  102yþ1

(102 )3 (103 )1=6 (e) pffiffiffiffiffi 10  (104 )1=2

(c)

31=2  32=3 31=2  31=3

( f ) ½(x1 )2 3

pffiffiffiffiffiffiffiffiffiffiffiffiffi 272=3 þ 52=3  51=3  0 1 þ 21  161=2  4  30 (b) 4 2 1 (c) 82=3 þ 32  (10)0 9

(a)

(d ) 272=3  3(3x)0 þ 251=2

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( f ) 3 (x  2)2

when x ¼ 6

41=2 a2=3 b1=6 c3=2 82=3 a1=3 b2=3 c5=2  8 4 1=4 2 3 (h) 54 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sp ffiffiffiffiffi ffiffiffiffiffip 3 4 a2 b5 (i) c2 d 2 (g)

(g) x3=2 þ 4x1  5x0

when

x¼4

(h) y2=3 þ 3y1  2y0

when

y ¼ 1=8

pffiffiffi (i) 642=3  165=4  20  ( 3)4 pffiffiffi 2=3 aa a5=6 p ffiffiffiffi ffi p ffiffiffiffi ffi ( j) þ 6 3 a2  a1=2 a5

(e) 82=3  163=4  20  82=3

7.22

(m) (x  y)0 ½(x  y)4 1=2

(k)

! pffiffiffiffiffiffiffiffiffiffiffi 72y2n 0  9 (2ynþ2 )1 3

(a) 250 þ 0:251=2  81=3  41=2 þ 0:0271=3

( f ) (64)2=3  3(150)0 þ 12(2)2

1  3a0 þ (3a)0 þ (27)1=3  13=2 82=3 32 þ 5(2)0 (c) 3  4(3)1 30 x þ 4x1 if x¼8 (d ) x2=3 1 2þ2 (e) þ (8)0  43=2 5

3 (g) (0:125)2=3 þ 2 þ 21 rffiffiffiffiffiffiffiffiffi n 32 (h) 25þn (60 000)3 (0:00 002)4 (i) 1002 (72 000 000)(0:0002)5

(b)

CHAP. 7]

7.23

57

EXPONENTS

(a)

(x2 þ 3x þ 4)1=3 ½ 12 (5  x)1=2   (5  x)1=2 ½(x2 þ 3x þ 4)2=3 (2x þ 3)=3 (x2 þ 3x þ 4)2=3

(b)

(9x2  5y)1=4 (2x)  x2 ½14 (9x2  5y)3=4 (18x) (9x2  5y)1=2

x ¼ 2,

if

x¼1

if

y¼4

(x þ 1)2=3 ½12 (x  1)1=2   (x  1)1=2 ½23 (x þ 1)1=3  (x þ 1)4=3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d ) x  1 þ x2 þ 2x þ 1

(c)

(e) 3x  2y 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4x2  4xy þ y2

ANSWERS TO SUPPLEMENTARY PROBLEMS 7.19

(a) 81

(d ) 1=64

(b) 8x3

(e)

(c)

27y3 64 9

(g)

1 16x2

( f ) y=2

7.20

(a) 2

(b) 1=10

7.21

(a)

16 3

(b)

7 2

(c) 4

7.22

(a) 0:8

(b)

4 3

(c)

7.23

(a) 

4x5 3y7

( j)

1 a2 b2

(m)

1 (x  y)2

(h) 2

(k) 3=2

(n) x5y

(i) 1=2

( l) 1

(o) 3y2

(c) 1

(d) 1 (d) 11

46 15

(d ) 34

4

(e) 10 (e)

1 4

(f )

(e) 

13 2

(f ) x 1 4 (f )

6

(g) 4 1 16

1 3

(b)

7 8

(c)

7x 6(x  1)1=2 (x þ 1)5=3

(d ) 2x if x  1, (e) x  y if 2x  y,

2 if 5x  3y

x  1 if 2x  y

( p) 7200

pffiffiffi a b (g) 8c4 89 (h) 4 (g)

26 5

(h)

4 15

(i) 18

(h)

1 2

a1=4 b5=6 c d pffiffiffi 2 2 ( j) (k) 2 y a (i)

(i) 150

CHAPTER 8

Radicals 8.1

RADICAL EXPRESSIONS

8.2

LAWS FOR RADICALS

pffiffiffi A radical is an expression of the form n a which denotes the principal nth root of a. The positive integer n is the index, or order, the radical and the number a is the radicand. The index is omitted if n ¼ 2: ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pof pffiffiffiffiffiffiffiffiffiffiffiffiffi Thus 3 5, 4 7x3  2y2 , x þ 10 are radicals which have respectively indices 3, 4, and 2 and radicands 5, 7x3  2y2 , and x þ 10:

The laws for radicals are the same as the laws for exponents, since frequently used. Note: If n is even, assume a, b  0: pffiffiffi A. ( n a)n ¼ a pffiffiffi ( 3 6)3 ¼ 6,

EXAMPLES 8.1.

B.

b=0 rffiffiffiffiffi p ffiffiffi p ffiffiffi 5 5 5 5 5 5 ffiffiffi ¼ , ¼p 5 2 32 32

EXAMPLES 8.3.

ffiffiffiffiffiffi p pffiffiffi n am ¼ ( n a) m

EXAMPLE 8.4.

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 (x þ 1)3 xþ1 3 (x þ 1) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ¼p 6 3 6 (y  2)2 (y  2) (y  2)

p ffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 3 (27)4 ¼ ( 3 27)4 ¼ 34 ¼ 81

p ffiffiffiffiffiffiffiffiffi p pffiffiffi m n a ¼ mn a

EXAMPLES 8.5.

8.3

ffiffiffiffiffiffiffiffiffi p p ffiffiffiffiffi pffiffiffiffiffi 7 2 5 x y ¼ 7 x2 7 y5

ffiffiffiffiffi p p pffiffiffi ffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi 3 54 ¼ 3 27  2 ¼ 3 27  3 2 ¼ 3 3 2,

rffiffiffi p ffiffiffi n a n a ffiffiffi C. ¼p n b b

E.

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( 4 x2 þ y2 )4 ¼ x2 þ y2

ffiffiffi p pffiffiffi pffiffiffi n ab ¼ n a n b

EXAMPLES 8.2.

D.

ffiffiffi p n a ¼ a1=n . The following are the laws most

ffiffiffiffiffiffiffiffiffi pffiffiffi p p 3 5 ¼ 6 5,

ffiffiffiffiffiffi p pffiffiffi pffiffiffi 4 3 2 ¼ 12 2,

ffiffiffiffiffiffiffiffi p pffiffiffiffiffi pffiffiffiffiffi 5 3 x2 ¼ 15 x2

SIMPLIFYING RADICALS

The form of a radical may be changed in the following ways. 58

CHAP. 8]

RADICALS

59

(a) Removal of perfect nth powers from the radicand. EXAMPLES 8.6.

ffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi p pffiffiffi ffiffiffiffiffi p 3 32 ¼ 3 23 (4) ¼ 3 23  3 4 ¼ 2 3 4 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 8x5 y7 ¼ (4x4 y6 )(2xy) ¼ 4x4 y6 2xy ¼ 2x2 y3 2xy

(b) Reduction of the index of the radical. ffiffiffiffiffi pffiffiffiffiffi pffiffiffi p ffiffiffiffiffi p pffiffiffi 4 4 6=4 3=2 6 EXAMPLES 8.7. ¼ffiffiffiffiffiffiffi 23 ¼pffiffiffi8 ¼ 2 2 where the index is reduced from 4 to 2. ffiffiffiffiffiffiffiffiffiffiffiffi 64 ¼ 2 ¼ 2 ¼ 2 p p ffiffiffiffiffiffiffiffiffi p 6 3 6 3 25x6 p x ffiffiffiffiffiffiffiffiffiffiffi 5, where the index is reduced ¼ffiffiffiffiffi(5x3 )2=6 ¼ (5x3 )1=3 ¼ 5x3 ¼ p ¼ffiffiffiffiffiffiffiffiffiffiffi (5x3 )2p pfrom ffiffiffiffiffiffiffi 6 to 3. Note. 4 (4)2 ¼ 4 16 ¼ 2. It is incorrect to write 4 (4)2 ¼ (4)2=4 ¼ (4)1=2 ¼ 4.

(c) Rationalization of the denominator in the radicand. pffiffiffiffiffiffiffiffi EXAMPLE 8.8. Rationalize the denominator of 3 9=2: Multiply the numerator and denominator of the radicand (9/2) by such a number as will make the denominator a perfect nth power (here n ¼ 3) and then remove the denominator from under the radical sign. The number in this case is 22. Then rffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffi  2  rffiffiffiffiffiffiffiffiffiffi 3 2 3 9(2 ) 36 2 3 9 3 9 : ¼ ¼ ¼ 2 2 2 22 23 rffiffiffiffiffiffiffiffiffiffiffi 3 2 4 7a y EXAMPLE 8.9. Rationalize the denominator of : 8b6 x3 6 3 To make 8b x a perfect 4th power, multiply numerator and denominator by 2b2 x. Then rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3 2 3 2 2b2 x 3 2 2 14a3 y2 b2 x 4 7a y 4 7a y 4 14a y b x  ¼ ¼ ¼ 2b2 x 8b6 x3 8b6 x3 2b2 x 16b8 x4

A radical is said to be in simplest form if: (a) all perfect nth powers have been removed from the radical, (b) the index of the radical is as small as possible, (c) no fractions are present in the radicand, i.e., the denominator has been rationalized.

8.4

OPERATIONS WITH RADICALS

Two or more radicals are said to be similar if after being reduced to simplest form they have the same index and radicand. pffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffi Thus 32, 1=2, and 8 are similar since rffiffiffi rffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi 2 1 1 2 32 ¼ 16  2 ¼ 4 2, , and 8 ¼ 4  2 ¼ 2 2:  ¼ ¼ 2 2 2 2 pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi Here each radicand is 2 and each index is 2. However, 3 32 and 3 2 are dissimilar since 3 32 ¼ 3 8  4 ¼ 2 3 4: To add algebraically two or more radicals, reduce each radical to simplest form and combine terms with similar radicals. Thus: pffiffiffi   pffiffiffi pffiffiffi 3 pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffi 2 1 2 2¼ 4 2 2¼ 32  1=2  8 ¼ 4 2  2: 2 2 2

60

RADICALS

[CHAP. 8

When multiplying two radicals, we choose the procedure to use based on whether or not the indices of the radicals are the same. (a)

To multiply two or more radicals having the same index, use Law B: ffiffiffi pffiffiffi ffiffiffi p p n n a b ¼ n ab: EXAMPLES 8.10.

(b)

pffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi (2 3 4)(3 3 16) ¼ 2  3 3 4 3 16 ¼ 6 3 64 ¼ 6  4 ¼ 24 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi (3 4 x2 y)( 4 x3 y2 ) ¼ 3 4 (x2 y)(x3 y2 ) ¼ 3 4 x5 y3 ¼ 3x 4 xy3 :

To multiply radicals with different indices it is convenient to use fractional exponents and the laws of exponents. EXAMPLES 8.11.

ffiffiffipffiffiffi p ffiffiffiffiffiffiffiffi p 3 6 1=6 5ffiffiffip2ffiffiffi ¼ 5p1=3  2ffiffi1=2 52=6  23=6 ¼ (52  23 )1=6 ¼ (25 p 8)ffiffiffiffi 200 ffiffiffiffiffip ffi ¼p p ffiffiffi . ffi ¼2=3 3 6 3 6 4 2 ¼ 22 2 ¼ 2  21=2 ¼ 24=6  23=6 ¼ 27=6 ¼ 27 ¼ 2 2

When dividing two radicals, we choose the procedure to use based on whether or not the indices of the radicals are the same. (a)

To divide two radicals having the same index, use Law C, p ffiffiffi rffiffiffi n a a ffiffiffi ¼ n , p n b b and simplify. EXAMPLE 8.12.

p ffiffiffi rffiffiffi rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi p ffiffiffi 3 3 2 5 45 3 5 3 3 5 3 45 p ffiffi ffi ¼ ¼  ¼ ¼ 3 3 3 3 32 33 3

We may also rationalize the denominator directly, as follows. ffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi p p ffiffiffi p ffiffiffi p ffiffiffiffiffi 3 3 3 3 5  32 5 5 3 32 45 ffiffi ffi ffiffi ffi ffiffiffiffi ffi ffiffiffiffi ffi p p p p ¼  ¼ ¼ 3 3 3 3 3 3 3 32 33

(b)

To divide two radicals with different indices it is convenient to use fractional exponents and the laws of exponents. EXAMPLES 8.13. rffiffiffiffiffi rffiffiffiffiffi pffiffiffi 2 pffiffiffiffiffi 6 61=2 62=4 4 6 4 36 p ffiffi ffi ¼ ¼ ¼ 4 18 ¼ ¼ 4 1=4 1=4 2 2 2 2 2 ffiffiffiffiffi p ffiffiffi p 3 3 pffiffiffi 22 22=3 24=6 4 pffiffiffi ¼ pffiffiffi ¼ 1=2 ¼ 3=6 ¼ 21=6 ¼ 6 2 2 2 2 2

8.5

RATIONALIZING BINOMIAL DENOMINATORS pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi The numbers a þ b and a  b are called conjugates of each other. Thus 2 3 þ p ffiffiffi binomial pffiffiffi pffiffiffi irrational 2 and 2 3  2 are conjugates. The property of these conjugates that makes them useful is the fact that they are the sum andpdifference two ffiffiffi ffiffiffi their product is the difference of the squares of ffiffiffi pffiffiffi pofffiffiffi thepsame pffiffiffi terms,pso these terms. Hence, ( a þ b)( a  b) ¼ ( a)2  ( b)2 ¼ a  b: To rationalize a fraction whose denominator is a binomial with radicals of index 2, multiply numerator and denominator by the conjugate.

CHAP. 8]

RADICALS

61

EXAMPLE 8.14. pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 5 5 2 3  2 5(2 3  2) 2 3  2 pffiffiffi pffiffiffi ¼ pffiffiffi pffiffiffi  pffiffiffi pffiffiffi ¼ ¼ 12  2 2 2 3þ 2 2 3þ 2 2 3 2

pffiffiffi pffiffiffi by If the denominator of a fraction is 3 a þ 3 b, we multiply the numerator and denominator p p ffiffiffiffi ffi ffiffiffiffi ffi ffiffiffiffiffi p ffiffiffi ffiffiffi thepfraction pof 3 3 original denominator has the form 3 a  3 b, we mula2  3 ab þ b2 and get a denominator of a þ b: If the ffiffiffiffi ffi ffiffiffiffi ffi p p p ffiffiffiffiffi tiply the numerator and denominator of the fraction by 3 a2 þ 3 ab þ 3 b2 and get a denominator of a  b. (See Section 4.2 for the special product rules.) EXAMPLE 8.15.

pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi 3 3( 3 25 þ 2 3 5 þ 4) 3( 3 25 þ 2 3 5 þ 4) ffiffi ffi ffiffi ffi ffiffi ffi p p p ffiffiffiffiffi p ffiffi ffi p ¼ ¼ 3 ( 3 5)3  (2)3 5  2 ( 3 5  2)( 3 25 þ 2 3 5 þ 4) pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi p p ffiffiffi ffiffiffiffiffi 3( 3 25 þ 2 3 5 þ 4) 3( 3 25 þ 2 3 5 þ 4) 3 3 ¼ ¼ ¼  25  2 5  4 58 3

Solved Problems Reduction of a Radical Expression to Simplest Form

8.1

(a) (b) (c) (g) (h) (i)

( j) (k)

(l ) (m) (n) (o) (p) (q)

pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 18 ¼ 9  2 ¼ 32  2 ¼ 3 2 (d) 3 648 ¼ 3 8  27  3 ¼ 3 23  33  3 ¼ 6 3 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi ffiffiffiffiffi p pffiffiffi 3 80 ¼ 3 8  10 ¼ 3 23  10 ¼ 2 3 10 (e) a 9b4 c3 ¼ a 32 b4 c2  c ¼ 3ab2 c c ffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi p pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 6 5 3 243 ¼ 5 3 27  9 ¼ 5 3 33  9 ¼ 15 3 9 ( f ) 6 343 ¼ 73 ¼ 73=6 ¼ 71=2 ¼ 7 ffiffiffiffiffiffiffiffiffiffi p p ffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 6 Note that a  0: See (k) below. 81a2 ¼ 6 34 a2 ¼ 34=6 a2=6 ¼ 32=3 a1=3 ¼ 6 9a 2 pffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffi ffi 4x 3 64x7 y6 ¼ 3 43 x6  xy6 ¼ 4x2 y2 3 x ¼ 2 3 x y ffiffiffiffiffiffiffiffiffiffi p 5 (72)4 ¼ (72)4=5 ¼ (8  9)4=5 ¼ (23  32 )4=5 ¼ 212=5  38=5 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi ¼ (22  22=5 )(3  33=5 ) ¼ 22  3 5 22  33 ¼ 12 5 108 pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi (7 3 4ab)2 ¼ 49(4ab)2=3 ¼ 49 3 16a2 b2 ¼ 98 3 2a2 b2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 3)2 ¼ 2a(a þ 3). The student is reminded that (a þ 3)2 is a positive 2a a2 þ 6a þ 9 ¼ 2a p(a ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi number or zero; hence (a þ 3)2 ¼ a þ 3 only if a þ 3  0. If values of a such that a þ 3 , 0 are pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi included, we must write (a þ 3)2 ¼ j a þ 3 j: pffiffiffi pffiffiffi x  25 ( x þ 5)( x  5) pffiffiffi pffiffiffi pffiffiffi ¼ ¼ x5 xþ5 xþ5 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi pffiffiffi 12x4  36x2 y2 þ 27y4 ¼ 3(4x4  12x2 y2 þ 9y4 ) ¼ 3(2x2  3y2 ) ¼ (2x2  3y2 ) 3 Note that this is valid only if 2x2  3y2 : See (k) above. ffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p n n an b2n c3nþ1 dnþ2 ¼ (an b2n c3nþ1 d nþ2 )1=n ¼ ab2 c3 c1=n dd 2=n ¼ ab2 c3 d cd 2 ffiffiffiffiffiffiffiffiffiffiffi p p ffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffiffiffi pffiffiffiffiffiffiffiffiffi 3 256 ¼ 3 16 ¼ 3 8  2 ¼ 2 3 2 p ffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffi 4 3 6ab2 ¼ ½(6ab2 )1=3 1=4 ¼ (6ab2 )1=12 ¼ 12 6ab2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 5 729 a3 ¼ 5 729a3=2 ¼ (36 a3=2 )1=5 ¼ 312=10 a3=10 ¼ 3 10 9a3

Change in Form of a Radical

8.2

Express as radicals of the 12th order. pffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffi (a) 3 5 ¼ 51=3 ¼ 54=12 ¼ 12 54 ¼ 12 625

62

RADICALS

pffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 12 ab ¼ (ab)1=2 ¼ (ab)6=12 ¼ 12 (ab)6 ¼ a6 b6 p ffiffiffiffiffiffi pffiffiffiffiffi 12 (c) 6 xn ¼ xn=6 ¼ x2n=12 ¼ x2n

(b)

8.3

Express in terms of radicals of least order. pffiffiffi pffiffiffi (a) 4 9 ¼ 91=4 ¼ (32 )1=4 ¼ 31=2 ¼ 3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi (b) 12 8x3 y6 ¼ 12 (2xy2 )3 ¼ (2xy2 )3=12 ¼ (2xy2 )1=4 ¼ 4 2xy2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (c) 8 a2 þ 2ab þ b2 ¼ 8 (a þ b)2 ¼ (a þ b)2=8 ¼ (a þ b)1=4 ¼ 4 a þ b

8.4

Convert into entire radicals, i.e., radicals having coefficient 1. pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi (a) 6 3 ¼ 36  3 ¼ 108 pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 4x2 3 y2 ¼ 3 (4x2 )3 y2 ¼ 3 64x6 y2 ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffi rffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  4 4 2y 3 4 2x 4 2y 2x 2y 4 16x 4 32x ¼   ¼ (c) ¼ y x y x x y4 y3 rffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi ab aþb (a  b)2 a þ b ab  ¼ ¼ (d ) aþb ab aþb (a þ b)2 a  b

8.5

Determine which of the following irrational numbers is the greater: pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi (a) 3 2, 4 3 (c) 2 5, 3 2 (b) 5, 3 11 SOLUTION pffiffiffi ffiffiffi p 4 1=12 (a) 3 2 ¼ 21=3 ¼ 24=12 ¼ (24 )1=12 3 ¼ 31=4 ¼ 33=12 ¼ (33 )1=12 ¼ (27)1=12 : ffiffiffi ; p pffiffiffi ¼ (16) 3 1=12 1=12 4 Since (27) . (16) , 3 . 2: pffiffiffi p ffiffiffiffiffi 3 3 1=6 (b) 5 ¼ 51=2 ¼ 53=6 ¼ ¼ (125)1=6 ; 11 ¼ (11)1=3 ¼ (11)2=6 ¼ (112 )1=6 ¼ (121)1=6 : pffiffi(5 ffi )p ffiffiffiffiffi 3 Since 125 . 121, 5 . 11: pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 3 2 ¼ 32  2 ¼ 18: (c) 2 5 ¼ p2ffiffi2ffi  5 ¼ pffiffiffi 20; Hence 2 5 . 3 2:

8.6

Rationalize the denominator. rffiffiffi rffiffiffiffiffiffiffiffiffi rffiffiffiffiffi 2 2 3 6 1 pffiffiffi  ¼ 6 ¼ ¼ (a) 2 3 3 3 3 3 pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ffiffiffiffiffi 3 3 3 62 3 3 62 3 3 36 1 p 3 ffiffiffi ¼ p ffiffiffi  p ffiffiffiffiffi ¼ p ffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ 36 (b) p 3 3 3 3 2 2 6 2 6 6 6 66

pffiffiffiffiffi ffiffiffiffiffi 3 3 62=3 3  62=3 3 3 62 1 p 3 ffiffiffi ¼ 1=3  2=3 ¼ 36 Another method: p ¼ ¼ 3 1 2 6 6 6 6 6 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi 3 3 ffiffiffiffiffiffiffiffiffi 3 p ffiffiffiffiffiffiffiffiffi 4 y(2x) 3x p 4 y(8x ) 4 y 4 4 3y ¼ (c) 3x ¼ ¼ 3x ¼ 3x 8x 8x3 y 2x 2x 2 2x(2x)3 (2x)4 rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ab ab aþb a2  b2 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  b2 ¼  (d ) ¼ ¼ a aþb aþb aþb (a þ b)2 a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffi p pffiffiffiffiffiffiffiffiffi 4xy2 4xy2 3 (2xy2 )2 4xy2 3 (2xy2 )2 3 2 y4 ¼ 2y 3 4x2 y ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffi p ¼  ¼ (e) p ¼ 2 4x 3 3 2xy2 2xy2 2xy2 3 (2xy2 )2

[CHAP. 8

CHAP. 8]

RADICALS

Addition and Subtraction of Similar Radicals

8.7

(a) (b) (c) (d) (e)

(f) (g)

(h)

(i)

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 18 þ 50  72 ¼ 9  2 þ 25  2  36  2 ¼ 3 2 þ 5 2  6 2 ¼ (3 þ 5  6) 2 ¼ 2 2 pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi 2 27  4 12 ¼ 2 9  3  4 4  3 ¼ 2  3 3  4  2 3 ¼ 6 3  8 3 ¼ 2 3 rffiffiffiffiffiffiffiffiffi   pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffi 4 3 2   2  4 3 ¼ 20 þ 3   8 3 ¼ 14 3 4 75 þ 3 4=3  2 48 ¼ 4  5 3 þ 3 3 3 rffiffiffiffiffiffiffiffiffiffiffi 3   ffiffiffi 5 p ffiffiffi p ffiffiffiffiffiffiffiffiffiffi ffi p p ffiffiffiffiffiffiffiffiffiffi ffi p ffiffiffiffiffiffiffi ffi ffiffiffiffiffiffiffi ffi p p ffiffiffiffiffiffiffiffiffiffi 1 2 1 3 3 3 3 2¼ 2 432  3 250 þ 3 1=32 ¼ 3 63  2  3 53  2 þ ¼ 6  5 þ  5 2 2 4 4 pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi p ffiffiffiffiffi pffiffiffiffiffi 3 þ 3 81  27 þ 5 3 3 ¼ 3 þ 3 27  3  9  3 þ 5 3 3 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ 3 þ 3 3 3  3 3 þ 5 3 3 ¼ 2 3 þ 8 3 3 pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2a 3 27x3 y þ 3b 3 8x3 y  6c 3 x3 y ¼ 6ax 3 y þ 6bx 3 y þ 6cx 3 y ¼ 6x(a þ b þ c) 3 y rffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi rffiffiffi rffiffiffi rffiffiffiffiffiffiffiffiffi rffiffiffi 2 3 1 2 3 3 2 1 6  þ4  2 þ4 5  5 ¼2 3 8 24 3 3 8 2 24 6   2 1 5 pffiffiffi 5 pffiffiffi þ4  6¼ 6 ¼ 3 4 12 4 rffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 3 5 2 3 5 pffiffiffi þ pffiffiffiffiffiffiffi  1:6 ¼  þ pffiffiffiffiffiffiffiffiffiffi  (0:16)(10) 2 2 1=10 0:1 2 pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi 1 pffiffiffiffiffi 10 þ 3 10  0:4 10 ¼ 3:1 10 ¼ 2 rffiffiffi rffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffi pffiffiffiffiffi rffiffiffi ab a b 4 a b b a 4  3  þ pffiffiffiffiffi  pffiffiffiffiffi 3 þ pffiffiffiffiffi ¼ 2 2 b a a b a a ab ab ab     p ffiffiffiffiffi p ffiffiffiffiffi p ffiffiffiffiffi 2 3 4 2 3 4 pffiffiffiffiffi 2a  3b þ 4 pffiffiffiffiffi ab  ab þ ab ¼  þ ab ¼ ab ¼ b a ab b a ab ab

Multiplication of Radicals

8.8

63

pffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi (a) (2 7)(3 5) ¼ (2  3) 7  5 ¼ 6 35 pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffi (b) (3 3 2)(5 3 6)(8 3 4) ¼ (3  5  8) 3 2  6  4 ¼ 120 3 48 ¼ 240 3 6 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffi (c) ( 3 18x2 )( 3 2x) ¼ 3 36x3 ¼ x 3 36 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 4 (d) ab1 c5  4 a3 b3 c1 ¼ 4 a4 b2 c4 ¼ a2 bc2 ¼ ac b pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi (e) 3  3 2 ¼ 31=2  21=3 ¼ 33=6  22=6 ¼ 6 33  22 ¼ 6 108 ffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi p 4 ( f ) ( 3 14)( 4 686) ¼ ( 3 7  2)( 73  2) ¼ (71=3  21=3 )(73=4  21=4 ) ¼ (74=12  24=12 )(79=12  23=12 ) pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi ¼ 7(71=12  27=12 ) ¼ 7 12 7  27 ¼ 7 12 896 pffiffiffipffiffiffi (g) ( 5 3 x)6 ¼ 56=2 x6=3 ¼ 53 x2 ¼ 125x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (h) ( 4  106 )( 8:1  103 )( 0:0016) ¼ ( 4  106 )( 81  102 )( 16  104 ) ¼ (2103 )(910)(4102 ) ¼ 72104 ¼ 0:0072 pffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi pffiffiffipffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (i) ( 6 þ 3)( 6  2 3) ¼ 6 6 þ 3 6 þ ( 6)(2 3) þ ( 3)(2 3) pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ 6 þ 18  2 18  2  3 ¼  18 ¼ 3 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi ( j) ( 5 þ 2)2 ¼ ( 5)2 þ 2( 5)( 2) þ ( 2)2 ¼ 5 þ 2 10 þ 2 ¼ 7 þ 2 10 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (k) (7 5  4 3)2 ¼ (7 5)2  2(7 5)(4 3) þ (4 3)2 pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ 72  5  2  7  4 15 þ 42  3 ¼ 245  56 15 þ 48 ¼ 293  56 15

64

RADICALS

[CHAP. 8

pffiffiffi pffiffiffi pffiffiffi (l) ( 3 þ 1)( 3  1) ¼ ( 3)2  (1)2 ¼ 3  1 ¼ 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (m) (2 3  5)(2 3 þ 5) ¼ (2 3)2  ( 5)2 ¼ 4  3  5 ¼ 12  5 ¼ 7 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (n) (2 5  3 2)(2 5 þ 3 2) ¼ (2 5)2  (3 2)2 ¼ 4  5  9  2 ¼ 20  18 ¼ 2 pffiffiffi pffiffiffi pffiffiffi (o) (2 þ 3 3)(2  3 3) ¼ 4  3 9 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi ( p) (3 2 þ 2 3 4)(3 2  2 3 4) ¼ (3 2)2  (2 3 4)2 ¼ 18  4 3 16 ¼ 18  8 3 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (q) (3 2  4 5)(2 3 þ 3 6) ¼ (3 2)(2 3)  (4 5)(2 3) þ (3 2)(3 6)  (4 5)(3 6) pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi ¼ 6 6  8 15 þ 9 12  12 30 ¼ 6 6  8 15 þ 18 3  12 30 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (r) ( x þ y  z)( x þ y þ z) ¼ x þ y  z2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (s) (2 x  1  x 2)(3 x  1 þ 2x 2) ¼ 6(x  1)  3x 2(x  1) þ 4x 2(x  1)  4x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 6(x  1) þ x 2(x  1)  4x2 8.9

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (a) ( 2 þ 3 þ 5)( 2 þ 3  5) pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ ½( 2 þ 3) þ 5½( 2 þ 3)  5 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ ( 2 þ 3) 2  ( 5) 2 ¼ 2 þ 2 6 þ 3  5 ¼ 2 6 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (b) (2 3 þ 3 2 þ 1)(2 3  3 2  1) ¼ ½2 3 þ (3 2 þ 1)½2 3  (3 2 þ 1) pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ (2 3)2  (3 2 þ 1)2 ¼ 12  (9  2 þ 6 2 þ 1) ¼ 7  6 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (c) ( 2 þ 3 þ 5)2 ¼ ( 2)2 þ ( 3)2 þ ( 5)2 þ 2( 2)( 3) þ 2( 3)( 5) þ 2( 2)( 5) pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ 2 þ 3 þ 5 þ 2 6 þ 2 15 þ 2 10 ¼ 10 þ 2 6 þ 2 15 þ 2 10 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (d ) 6þ3 3  6  3 3 ¼ (6 þ 3 3)(6  3 3) ¼ 36  9  3 ¼ 9 ¼ 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (e) ( a þ b  a  b)2 ¼ a þ b  2 (a þ b)(a  b) þ a  b ¼ 2a  2 a2  b2

Division of Radicals. Rationalization of Denominators

8.10

(a) (b) (c) (d ) (e) (f) (g)

rffiffiffi pffiffiffi pffiffiffi 10 6 10 6 pffiffiffi ¼ ¼2 3 5 2 5 2 rffiffiffiffiffi pffiffiffiffiffi ffiffiffi 2 4 30 2 4 30 2 p 4 p ffiffiffi ¼ 6 ¼ 4 3 5 3 3 5 s ffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffi 4x 3 x2 y2 4x 3 x2 y2 4x p ffiffiffiffiffi 3  p xy ¼ ffiffiffiffiffi ¼ 3 xy y y y xy pffiffiffi rffiffiffi rffiffiffiffiffiffiffiffiffi rffiffiffi 2 2 3 6 1 pffiffiffi 2 pffiffiffi ¼  ¼ ¼ ¼ 6 3 3 3 9 3 3 p ffiffiffi rffiffiffi rffiffiffiffiffiffiffiffiffi rffiffiffiffiffi 3 ffiffiffiffiffi 2 1p 3 18 3 2 3 2 9 3 p ffiffi ffi  18 ¼ ¼ ¼ ¼ 3 3 3 9 27 3 3 rffiffiffi rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi ffiffiffiffiffi 1p 5 1 5 1 16 5 16 5  16 ¼ ¼ ¼ 2 2 16 32 2 pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 6 þ 4  2  5 16 6  12 3þ4 25 8 3þ4 25 8 pffiffiffi pffiffiffi ¼ ¼  pffiffiffi ¼ 2 2 2 2 2

CHAP. 8]

(h) (i) ( j) (k) (l ) (m)

RADICALS

pffiffiffi pffiffiffi pffiffiffi pffiffiffi 5  2 3( 5  2) pffiffiffi pffiffiffi 3 3 pffiffiffi pffiffiffi ¼ pffiffiffi pffiffiffi  pffiffiffi pffiffiffi ¼ ¼ 5 2 52 5 2 5þ 2 5þ 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 1þ 2 1þ 2 1þ 2 1þ2 2þ2 pffiffiffi ¼ pffiffiffi  pffiffiffi ¼ ¼ (3 þ 2 2) 12 1 2 1 2 1þ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 (x þ x2  y2 )  (x  x2  y2 ) 2 x2  y2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ y2 x  x2  y2 x þ x2  y2 (x  x2  y2 )(x þ x2  y2 ) ffiffiffiffiffiffiffiffiffiffiffiffi ffi p ffiffiffi pffiffiffi pffiffiffi p 6 3 ffiffiffiffiffiffiffiffi 3 3 3 3 4 3(33=6 )(42=6 ) 3 33  42 3 p ffiffiffi ffiffiffi ¼ p p ffiffiffi ¼ p ffiffiffi  p ¼ 6 432 ¼ 3 3 3 3 8 8 4 8 4 2 4 2 4 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi x1 xþ1 x  1  x þ 1 (x  1)  2 (x  1)(x þ 1) þ (x þ 1) pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ ¼ x2  1  x (x  1)  (x þ 1) x1þ xþ1 x1 xþ1 pffiffiffi pffiffiffi pffiffiffi p ffiffiffi pffiffiffi xþ x xþ x 1þx x x2 þ x x2 þ x pffiffiffi pffiffiffi  pffiffiffi ¼ ¼ ¼ 1 þ x þ x 1 þ x þ x 1 þ x  x (1 þ x)2  x 1 þ x þ x2

1 1 ffiffiffi ¼ 1=3 ffiffiffi p (n) p Let x ¼ 31=3 , y ¼ 41=3 . Then 3 3 3 þ 41=3 3þ 4 pffiffiffi ffiffiffiffiffi p ffiffiffi p 3 1 x2  xy þ y2 x2  xy þ y2 32=3  31=3 41=3 þ 42=3 9  3 12 þ 2 3 2  ¼ ¼ ¼ x þ y x2  xy þ y2 7 x3 þ y3 (31=3 )3 þ (41=3 )3 p p p ffiffiffiffi ffi p ffiffiffiffi ffi ffiffiffiffi ffi ffiffiffiffi ffi ffiffiffiffi ffi ffiffiffiffi ffi p p 3 x2 þ 3 xy þ 3 y2 1( 3 x2 þ 3 xy þ 3 y2 ) 1 ffiffi ffi p ffiffiffiffi ffi ¼ (o) p ¼ p ffiffiffiffi ffi p ffiffi ffi ffiffi ffi p ffiffiffiffi ffi ffiffi ffi p p 3 xy x  3 y ( 3 x  3 y)( 3 x2 þ 3 xy þ 3 y2 ) p p p ffiffiffiffiffi ffi ffiffiffiffiffiffi pffiffiffiffiffiffi p ffiffiffiffi ffi ffiffiffiffiffi pffiffiffiffiffiffi 3 3 3 3 mþn (m þ n)( m2  3 mn þ n2 ) (m þ n)( m2  3 mn þ n2 ) ffiffiffiffi p ffiffiffi ¼ pffiffiffiffi pffiffiffi pffiffiffiffiffiffi pffiffiffiffiffiffi pffiffiffiffiffi ¼ (p) p 3 (m þ n) m þ 3 n ( 3 m þ 3 n)( 3 m2  3 mn þ 3 n2 ) p ffiffiffiffiffiffi pffiffiffiffiffiffi p ffiffiffiffiffi 3 3 ¼ m2  3 mn þ n2

Supplementary Problems Show that:

8.11

pffiffiffiffiffi pffiffiffi (a) 72 ¼ 6 2 pffiffiffiffiffi pffiffiffi 27 ¼ 3 3 pffiffiffi pffiffiffiffiffi (c) 3 20 ¼ 6 5

(b)

pffiffiffi 2 pffiffiffiffiffiffiffiffiffi2ffi 50a ¼ 2a 2 (assuming a  0) (d ) 5 pffiffiffiffiffi a pffiffiffiffiffiffiffiffiffiffiffiffiffiffi (e) 75a3 b2 ¼ 5a2 3a b

8.12

pffiffiffiffiffi ab b pffiffiffiffiffi pffiffiffiffiffiffiffiffi ( j) 14 2=7 ¼ 2 14 pffiffiffiffiffiffiffiffi (i) a=b ¼

pffiffiffiffiffiffiffiffi pffiffiffiffiffi (k) 3 3 2=3 ¼ 3 18

(f)

pffiffiffiffiffi 4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 98a2 b3 ¼ 28 2b ab pffiffiffiffiffiffiffiffi pffiffiffiffiffi (g) 3 640 ¼ 4 3 10 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi (h) 3 88x3 y6 z5 ¼ 2xy2 z 3 11z2

rffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffi 3a 3 3 3p 3 ¼ 12a2 4 2a 8 sffiffiffiffiffiffiffiffiffiffiffi 5 1 pffiffiffiffiffiffiffiffiffi (m) xyz ¼ 10yz 2 2x yz 2 pffiffiffi pffiffiffiffiffiffiffiffiffiffi (n) 60 4=45 ¼ 8 5 pffiffiffiffiffiffiffiffi pffiffiffi (o) 3 4 4=9 ¼ 6

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffi 27 þ 48  12 ¼ 5 3 pffiffiffi pffiffiffiffiffi pffiffiffi (b) 5 8 3 18 ¼ 2 pffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffi (c) 2 150  4 54 þ 6 48 ¼ 24 3  2 6

pffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffi (d ) 5 2  3 50 þ 7 288 ¼ 74 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi (a  0) (e) 16a3 48a2 b ¼ 4a a3b p ffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffi p p ffiffiffi 3 3 3 ( f ) 3 16a3 þ 8 a3 =4 ¼ 10a 2

(a)

(l)

65

66

RADICALS pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffiffiffiffiffiffi p 3 128 ¼ 2 6 2 (k) ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p p ffiffiffiffiffiffiffi n nþ1 2n1 3n z ¼ xy2 z3 n x=y (l) (h) x y ffiffiffiffiffi p p ffiffi ffi ffiffi ffi p (i) 3 4 9  2 6 27 ¼ 3 (m) pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffi ( j) 6 8a3 =3  2 24ab2 þ a 54a ¼ (7a  4b) 6a (g)

8.13

pffiffiffi pffiffiffi pffiffiffiffiffi (a) (3 8)(6 5) ¼ 36 10

(l)

(b)

pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 48x5 3x3 ¼ 12x4

(m)

(c)

p ffiffiffi p ffiffiffiffiffi 3 2 3 32 ¼ 4

(n)

(d )

pffiffiffi pffiffiffi pffiffiffiffiffi 2( 2 þ 18) ¼ 8

(o)

(e) (5 þ

pffiffiffi pffiffiffi 2)(5  2) ¼ 23

(p)

( f ) (x þ

pffiffiffi pffiffiffi y)(x þ y) ¼ x2  y

(q)

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (g) (2 3  6)(3 3 þ 3 6) ¼ 9 2

pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi (x þ 1) 3 16x2  4x 3 x2 =4 ¼ 2 3 2x2 pffiffiffiffiffi pffiffiffiffiffiffiffiffi pffiffiffiffiffi 2 54  6 2=3  96 ¼ 0 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi ffi 3 ffi  5 6 y3 =x3 ¼ 4x  5y þ 3 pffiffiffiffi ffiffiffiffiffiffiffiffi xy 4 x=y þ p 4 2 2 xy x y a,b  0

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 82 7 8þ2 7¼6 pffiffiffi pffiffiffi 4þ 8 ¼2þ 2 2 pffiffiffiffiffi pffiffiffi 6  18 ¼2 2 3 rffiffiffi 3 1 pffiffiffi pffiffiffi  ¼ 2 2 2 pffiffiffiffiffi pffiffiffi 8 þ 4 48 ¼1þ2 3 8 pffiffiffiffiffi pffiffiffi 36  2 3 81 ¼6 3 3 6

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (h) (3 2  2 3)(4 2 þ 3 3) ¼ 6 þ 6 pffiffiffi pffiffiffi pffiffiffi pffiffiffi (i) ( 2  3)2 þ ( 2 þ 3)2 ¼ 10 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ( j) (2 a þ 5 a  b)( a þ a  b) ¼ 7a  5b þ 7 a2  ab pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (k) ( 3 þ 5 þ 7)( 3 þ 5  7) ¼ 1 þ 2 15

8.14

(a)

(b)

(c)

(d )

(e)

pffiffiffiffiffiffiffiffiffi pffiffiffi 2 24x3 pffiffiffiffiffi ¼ 4x 2 (x . 0) 3x pffiffiffi pffiffiffiffiffi a b ab pffiffiffi ¼ b b a pffiffiffi pffiffiffi 3þ 2 1 pffiffiffi pffiffiffi ¼1þ 6 2 2 pffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi 6  10  12 3 5 6 pffiffiffiffiffi pffiffiffi ¼ 18 3 rffiffiffiffiffiffiffiffiffiffi 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 9V 1 p 3 36pV 2 ¼ 16p2 4p

pffiffiffiffiffi 6a ffiffiffiffiffi ¼ a 3 18 ( f) p 3 12 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 ffiffiffiffiffiffiffi 3a7 b6 c5 ab p 5 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 4c4 (g) p 5 2 24a2 bc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 y3 z1 ffiffiffiffiffiffiffi 1 p 3 x 3 (h) ¼ 2y2 2 2 4xyz 2xy z

pffiffiffi pffiffiffip ffiffiffiffiffiffiffiffi ffiffiffi p 6 2 239 648 p ffiffi ffi ¼ (i) 3 ¼ 3 3 3 p ffiffiffiffiffi p ffiffiffiffiffi 3 ffiffiffiffiffi 1 p ffiffiffiffiffi 20  3 18 1 p 3 3 p ffiffiffiffiffi 45  12 ¼ ( j) 3 3 2 12 pffiffiffi 7þ2 1 ¼ (k) pffiffiffi 3 72 (l )

pffiffiffi 5 5 pffiffiffi ¼ (3  2) 3þ 2 7

pffiffiffi 2 pffiffiffi ¼ 4  2 3 2 3 pffiffiffi pffiffiffi s 3 3s s 3 (n) pffiffiffi ¼ þ 2 2 31 pffiffiffi pffiffiffi 2 31 (o) pffiffiffi ¼5 38 3þ2 pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 1 xþ1 2 xþ1x2 pffiffiffiffiffiffiffiffiffiffiffi ¼ (p) x 1þ xþ1 (m)

[CHAP. 8

CHAPTER 9

Simple Operations with Complex Numbers 9.1

COMPLEX NUMBERS

pffiffiffiffiffiffiffi The unit of imaginary numbers is 1 and is designated by the letter i. Many laws which hold for real numbers hold for imaginary pffiffiffiffiffiffiffi pnumbers pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffipffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi as well. Thus 4 ¼ (4)(1) ¼ 2 1 ¼ 2i, 18 ¼ (18)(1) ¼ 18 1 ¼ 3 2i. Also, since i ¼ 1, we have i2 ¼ 1, i3 ¼ i2  i ¼ (1)i ¼ i, i4 ¼ (i2 )2 ¼ (1)2 ¼ 1, i5 ¼ i4  i ¼ 1  i ¼ i, and similarly for any integral power of i. Note. One must be very careful in applying some of the laws which hold for real numbers. For example, one might be tempted to write pffiffiffiffiffiffiffipffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 4 4 ¼ (4)(4) ¼ 16 ¼ 4, which is incorrect: To avoid such difficulties, always express ever it arises. Thus:

pffiffiffiffiffiffiffiffi pffiffiffiffi m, where m is a positive number, as mi; and use i2 ¼ 1 when-

pffiffiffiffiffiffiffipffiffiffiffiffiffiffi 4 4 ¼ (2i)(2i) ¼ 4i2 ¼ 4,

which is correct:

pffiffiffiffiffiffiffi A complex number is an expression of the form a þ bi, where a and b are real numbers and i ¼ 1. In the complex number a þ bi, a is called the real part and bi the imaginary part. When a ¼ 0, the complex number is called a pure imaginary. If b ¼ 0, the complex number reduces to the real number a. Thus complex numbers include all real numbers and all pure imaginary numbers. Two complex numbers a þ bi and c þ di are equal if and only if a ¼ c and b ¼ d. Thus a þ bi ¼ 0 if and only if a ¼ 0, b ¼ 0. If c þ di ¼ 3, then c ¼ 3, d ¼ 0. The conjugate of a complex number a þ bi is a  bi, and conversely. Thus 5  3i and 5 þ 3i are conjugates.

9.2

GRAPHICAL REPRESENTATION OF COMPLEX NUMBERS

Employing rectangular coordinate axes, the complex number x þ yi is represented by, or corresponds to, the point whose coordinates are (x, y). See Fig. 9-1. 67

68

SIMPLE OPERATIONS WITH COMPLEX NUMBERS

[CHAP. 9

Fig. 9-1

To represent the complex number 3 þ 4i, measure off 3 units distance along X 0 X and to the right of O, and then up 4 units distance. To represent the number 2 þ 3i, measure off 2 units distance along X 0 X and to the left of O, and then up 3 units distance. To represent the number 1  4i, measure off 1 unit distance along X 0 X and to the left of O, and then down 4 units distance. To represent the number 2  4i, measure off 2 units distance along X 0 X and to the right of O, and then down 4 units distance. Pure imaginary numbers (such as 2i and 2i) are represented by points on the line Y 0 Y. Real numbers (such as 4 and 23) are represented by points on the line X 0 X.

9.3

ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS

To add two complex numbers, add the real parts and the imaginary parts separately. Thus (a þ bi) þ (c þ di) ¼ (a þ c) þ (b þ d)i (5 þ 4i) þ (3 þ 2i) ¼ (5 þ 3) þ (4 þ 2)i ¼ 8 þ 6i (6 þ 2i) þ (4  5i) ¼ (6 þ 4) þ (2  5)i ¼ 2  3i: To subtract two complex numbers, subtract the real parts and the imaginary parts separately. Thus (a þ bi)  (c þ di) ¼ (a  c) þ (b  d)i (3 þ 2i)  (5  3i) ¼ (3  5) þ (2 þ 3)i ¼ 2 þ 5i (1 þ i)  (3 þ 2i) ¼ (1 þ 3) þ (1  2)i ¼ 2  i: To multiply two complex numbers, treat the numbers as ordinary binomials and replace i2 by 1. Thus (a þ bi)(c þ di) ¼ ac þ adi þ bci þ bdi2 ¼ (ac  bd) þ (ad þ bc)i (5 þ 3i)(2  2i) ¼ 10  10i þ 6i  6i2 ¼ 10  4i  6(1) ¼ 16  4i:

CHAP. 9]

SIMPLE OPERATIONS WITH COMPLEX NUMBERS

69

To divide two complex numbers, multiply the numerator and denominator of the fraction by the conjugate of the denominator, replacing i2 by 1. Thus 2þi 2 þ i 3 þ 4i 6 þ 8i þ 3i þ 4i2 2 þ 11i 2 11 ¼  ¼ ¼ þ i: ¼ 2 3  4i 3  4i 3 þ 4i 25 25 25 9  16i

Solved Problems 9.1

9.2

Express each of the following in terms of i. pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffipffiffiffiffiffiffiffi (a) 25 ¼ (25)(1) ¼ 25 1 ¼ 5i pffiffiffiffiffiffiffiffiffi pffiffiffiffiffipffiffiffiffiffiffiffi (b) 3 36 ¼ 3 36 1 ¼ 3  6  i ¼ 18i pffiffiffiffiffiffiffiffiffi pffiffiffiffiffipffiffiffiffiffiffiffi (c) 4 81 ¼ 4 81 1 ¼ 4  9  i ¼ 36i rffiffiffiffiffiffiffi rffiffiffi rffiffiffi pffiffiffi 2 1 1 pffiffiffiffiffiffiffi 2 1 ¼ (d )  ¼ i¼ i 2 2 2 4 rffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffi 16 49 4 7 8 21 16 21 1 (e) 2 3 ¼2 i3 i¼ i i¼ i i¼ i 25 100 5 10 5 10 10 10 2 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (f) 12  3 ¼ 12i  3i ¼ 2 3i  3i ¼ 3i pffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (g) 3 50 þ 5 18  6 200 ¼ 15 2i þ 15 2i  60 2i ¼ 30 2i pffiffiffiffiffiffiffi pffiffiffi (h) 2 þ 4 ¼ 2 þ 4i ¼ 2 þ 2i pffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffi (i) 6  50 ¼ 6  50i ¼ 6  5 2i pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffi pffiffiffi pffiffiffi ( j) 8 þ 8 ¼ 8 þ 8i ¼ 2 2 þ 2 2i pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 1 1 (10 þ 125) ¼ (10 þ 5 5i) ¼ 2 þ 5i (k) 5 5 pffiffiffi pffiffiffi pffiffiffi 1 pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffi ( 32 þ 128) ¼ (4 2 þ 8 2i) ¼ 2 þ 2 2i (l ) 4 4 pffiffiffiffiffiffiffi pffiffiffi p ffiffiffiffi 3 pffiffiffi 8 þ 8 2 þ 2 2i ¼ 1 þ 2i ¼ (m) 2 2 Perform the indicated operations both algebraically and graphically: (a) (2 þ 6i) þ (5 þ 3i), (b) (4 þ 2i)  (3 þ 5i):

Fig. 9-2

Fig. 9-3

70

SIMPLE OPERATIONS WITH COMPLEX NUMBERS

[CHAP. 9

SOLUTION

9.3

(a)

Algebraically: (2 þ 6i) þ (5 þ 3i) ¼ 7 þ 9i Graphically: Represent the two complex numbers by the points P1 and P2 respectively, as shown in Fig. 9-2. Connect P1 and P2 with the origin O. Complete the parallelogram having OP1 and OP2 as adjacent sides. The vertex P (point 7 þ 9i) represents the sum of the two given complex numbers.

(b)

Algebraically: (4 þ 2i)  (3 þ 5i) ¼ 7  3i Graphically: (4 þ 2i)  (3 þ 5i) ¼ (4 þ 2i) þ (3  5i). We now proceed to add (4 þ 2i) with (3  5i). Represent the two complex numbers (4 þ 2i) and (3  5i) by the points P1 and P2 respectively, as shown in Fig. 9-3. Connect P1 and P2 with the origin O. Complete the parallelogram having OP1 and OP2 as adjacent sides. The vertex P (point 7  3i) represents the subtraction (4 þ 2i)  (3 þ 5i).

Perform the indicated operations and simplify. (a) (5  2i) þ (6 þ 3i) ¼ 11 þ i (b) (6 þ 3i)  (4  2i) ¼ 6 þ 3i  4 þ 2i ¼ 2 þ 5i (c) (5  3i)  (2 þ 5i) ¼ 5  3i þ 2  5i ¼ 7  8i       3 5 1 1 3 1 5 1 5 7 þ i þ  þ i ¼  þ þ i¼ þ i (d ) 2 8 4 4 2 4 8 4 4 8 (e) (a þ bi) þ (a  bi) ¼ 2a ( f ) (a þ bi)  (a  bi) ¼ a þ bi  a þ bi ¼ 2bi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi (g) (5  125)  (4  20) ¼ (5  5 5i)  (4  2 5i) ¼ 1  3 5i

9.4

pffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffipffiffiffiffiffi pffiffiffiffiffi 2 32 ¼ ( 2i)( 32i) ¼ 2 32i2 ¼ 64(1) ¼ 8 pffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffipffiffiffiffiffi pffiffiffiffiffiffiffiffi (b) 3 5 20 ¼ 3( 5i)( 20i) ¼ 3( 5 20)i2 ¼ 3 100(1) ¼ 30

(a)

(c) (4i)(3i) ¼ 12i2 ¼ 12 (d ) (6i)2 ¼ 36i2 ¼ 36 pffiffiffiffiffiffiffi (e) (2 1)3 ¼ (2i)3 ¼ 8i3 ¼ 8i(i2 ) ¼ 8i ( f ) 3i(i þ 2) ¼ 3i2 þ 6i ¼ 3 þ 6i (g) (3  2i)(4 þ i) ¼ 3  4 þ 3  i  (2i)4  (2i)i ¼ 12 þ 3i  8i þ 2 ¼ 14  5i (h) (5  3i)(i þ 2) ¼ 5i þ 10  3i2  6i ¼ 5i þ 10 þ 3  6i ¼ 13  i (i)

(5 þ 3i)2 ¼ 52 þ 2(5)3i þ (3i)2 ¼ 25 þ 30i þ 9i2 ¼ 16 þ 30i

( j ) (2  i)(3 þ 2i)(1  4i) ¼ (6 þ 4i  3i  2i2 )(1  4i) ¼ (8 þ i)(1  4i) ¼ 8  32i þ i  4i2 ¼ 12  31i pffiffiffi pffiffiffi 2 pffiffiffi2 pffiffiffipffiffiffi  pffiffiffi 2 2 2 2 2 2 2 1 1 þ i ¼ i þ i ¼ þi ¼i (k) þ2 2 2 2 2 2 2 2 2 (l ) (1 þ i)3 ¼ 1 þ 3i þ 3i2 þ i3 ¼ 1 þ 3i  3  i ¼ 2 þ 2i (m) (3  2i)3 ¼ 33 þ 3(32 )(2i) þ 3(3)(2i)2 þ (2i)3 ¼ 27 þ 3(9)(2i) þ 3(3)(4i2 )  8i3 ¼ 27  54i  36 þ 8i ¼ 9  46i (n) (3 þ 2i)3 ¼ 33 þ 3(32 )(2i) þ 3(3)(2i)2 þ (2i)3 ¼ 27 þ 54i þ 36i2 þ 8i3 ¼ 27 þ 54i  36  8i ¼ 9 þ 46i (o) (1 þ 2i)4 ¼ ½(1 þ 2i)2 2 ¼ (1 þ 4i þ 4i2 )2 ¼ (3 þ 4i)2 ¼ 9  24i þ 16i2 ¼ 7  24i ( p) (1 þ i)8 ¼ ½(1 þ i)2 4 ¼ (1  2i þ i2 )4 ¼ (2i)4 ¼ 16i4 ¼ 16

CHAP. 9]

9.5

SIMPLE OPERATIONS WITH COMPLEX NUMBERS

1 þ i 1 þ i 3 þ i 3 þ 3i þ i þ i2 2 þ 4i 1 2 ¼  ¼ ¼ þ i ¼ 3i 3i 3þi 10 5 5 32  i 2   1 1 i i i ¼ ¼ i (b) ¼ 2¼ i i i i 1 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 3 þ 2i 2 3 þ 2i 3 2 þ 4 3i (2 3 þ 2i)(3 2 þ 4 3i) pffiffiffi pffiffiffi pffiffiffi ¼ pffiffiffi pffiffiffi  pffiffiffi pffiffiffi ¼ (c) pffiffiffi 3 2  4 3i 3 2  4 3i 3 2 þ 4 3i (3 2)2  (4 3)2 i2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 6 5 6 6 þ 8 9i þ 3 4i þ 4 6i2 2 6 þ 30i þ i ¼ ¼ ¼ 33 11 66 18 þ 48 (a)

Supplementary Problems pffiffiffiffiffiffiffi 4 þ 4 (g) 2 pffiffiffiffiffiffiffiffiffiffiffi 1 (h) (12  288) 6

9.6

Express each of the following in terms of i. pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (a) 2 49 (d ) 4 1=8 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (b) 4 64 (e) 3 25  5 100 pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (c) 6 1=9 ( f ) 2 72 þ 3 32

9.7

Perform the indicated operations both algebraically and graphically. (a) (3 þ 2i) þ (2 þ 3i) (c) (4  3i)  (2 þ i) (b) (2  i) þ (4 þ 5i)

9.8

(d ) (2 þ 2i)  (2  i)

Perform each of the indicated operations and simplify.

(a) (3 þ 4i) þ (1  6i) (b) (2 þ 5i)  (3  2i)     2 1 1 1  i   þ i (c) 3 2 3 2 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi (d ) (3 þ 8)  (2  32) pffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffi (e) 3 12 pffiffiffi pffiffiffi ( f ) (i 2)(i 2)

9.9

pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi (i) 4 81  3 36 þ 4 25 pffiffiffiffiffi pffiffiffiffiffiffiffiffiffi ( j) 3 12  3 12

(a)

2  5i 4 þ 3i

pffiffiffiffiffiffiffi (h) (12 3)6

1 2  2i pffiffiffi pffiffiffi 3 2 þ 2 3i pffiffiffi (c) pffiffiffi 3 2  2 3i

(o) (i  1)3

( j) (2 þ i)(2  i)

( p) (2 þ 3i)3

(k) (3 þ 4i)(3  4i)

(q) (1  i)4

(l ) (2  5i)(3 þ 2i)

(r)

pffiffiffi 3  2i pffiffiffi 2i 1 3 (e) þ 1  2i 1 þ 4i (f)

(n) (1 þ i)(2 þ 2i)(3  i)

(i) 5i(2  i)

(d )

(b)

(m) (3  4i)2

(g) (2i)4

5 10 þ 3  4i 4 þ 3i

(g)

i þ i2 þ i3 þ i4 1þi

i26  i i1  11 2 4i  i (i) 1 þ 2i

(h)

ANSWERS TO SUPPLEMENTARY PROBLEMS 9.6

(a) 14i (b) 32i

(c) 2i pffiffiffi (d ) 2i

9.7

(a) 5 þ 5i and Fig. 9-4 (b) 2 þ 4i and Fig. 9-5

(e) 35i (g) 2 þ i pffiffiffi pffiffiffi ( f ) 24 2i (h) 2  2 2i

(i þ 2)5

(i) 18i þ 20 pffiffiffi pffiffiffi ( j) 6 3  6 3i

(c) 6  4i and Fig. 9-6 (d ) 3i and Fig. 9-7

71

72

SIMPLE OPERATIONS WITH COMPLEX NUMBERS

Fig. 9-5

Fig. 9-4

Fig. 9-6

9.8

9.9

[CHAP. 9

Fig. 9-7

(a) 2  2i

pffiffiffi (d ) 1 þ 6 2i

(g) 16

( j) 5

(m) 7  24i

( p) 46 þ 9i

(b) 5 þ 7i (c) 1  i

(e) 6 (f) 2

(h) 27=64 (i) 5 þ 10i

(k) 25 (l ) 16  11i

(n) 4 þ 12i (o) 2 þ 2i

(q) 4 (r) 38 þ 41i

(a) 

7 26  i 25 25

1 1 (b)   i 4 4

1 2pffiffiffi 6i þ 5 5 3pffiffiffi (d ) 1  2i 2 (c)

(e)

32 26  i 85 85

(g) 0

(f )

11 2  i 5 5

(h) i

(i) 3 þ 4i

CHAPTER 10

Equations in General 10.1

EQUATIONS

An equation is a statement of equality between two expressions called members. An equation which is true for only certain values of the variables (sometimes called unknowns) involved is called a conditional equation or simply an equation. An equation which is true for all permissible values of the variables (or unknowns) involved is called an identity. By permissible values are meant the values for which the members are defined. EXAMPLE 10.1.

x þ 5 ¼ 8 is true only for x ¼ 3; it is a conditional equation.

EXAMPLE 10.2.

x2  y2 ¼ (x  y)(x þ y) is true for all values of x and y; it is an identity.

EXAMPLE 10.3. 1 1 2x  5 þ ¼ x  2 x  3 (x  2)(x  3) is true for all values except for the nonpermissible values x ¼ 2, x ¼ 3; these excluded values lead to division by zero which is not allowed. Since the equation is true for all permissible values of x, it is an identity. The symbol ; is sometimes used for identities instead of ¼ .

The solutions of a conditional equation are those values of the unknowns which make both members equal. These solutions are said to satisfy the equation. If only one unknown is involved the solutions are also called roots. To solve an equation means to find all of the solutions. Thus x ¼ 2 is a solution or root of 2x þ 3 ¼ 7, since if we substitute x ¼ 2 into the equation we obtain 2(2) þ 3 ¼ 7 and both members are equal, i.e., the equation is satisfied. Similarly, three (of the many) solutions of 2x þ y ¼ 4 are: x ¼ 0, y ¼ 4; x ¼ 1, y ¼ 2; x ¼ 5, y ¼ 6.

10.2

OPERATIONS USED IN TRANSFORMING EQUATIONS

A. If equals are added to equals, the results are equal. Thus if x  y ¼ z, we may add y to both members and obtain x ¼ y þ z. B. If equals are subtracted from equals, the results are equal. Thus if x þ 2 ¼ 5, we may subtract 2 from both members to obtain x ¼ 3. Note. Because of rules A and B we may transpose a term from one member of an equation to the other 73

74

C.

D.

E. F.

EQUATIONS IN GENERAL

[CHAP. 10

member merely by changing the sign of the term. Thus if 3x þ 2y  5 ¼ x  3y þ 2, then 3x  x þ 2y þ 3y ¼ 5 þ 2 or 2x þ 5y ¼ 7: If equals are multiplied by equals, the results are equal. Thus if both members of 14 y ¼ 2x2 are multiplied by 4 the result is y ¼ 8x2 : Similarly, if both members of 95 C ¼ F  32 are multiplied by 59 the result is C ¼ 59 (F  32). If equals are divided by equals, the results are equal provided there is no division by zero. Thus if 4x ¼ 12, we may divide both members by 4 to obtain x ¼ 3. Similarly, if E ¼ IR we may divide both sides by R = 0 to obtain I ¼ E/R. The same powers p offfiffiffiffiffiffi equals are equal. pffiffiffiffiffiffiffi ffi Thus if T ¼ 2p l=g, then T 2 ¼ (2p l=g)2 ¼ 4p2 l=g: The same roots of equals are equal. Thus: rffiffiffiffiffiffi 3V 3 3V 3 , then r¼ if r ¼ : 4p 4p

G. Reciprocals of equals are equal provided the reciprocal of zero does not occur. Thus if 1=x ¼ 1=3, then x ¼ 3. Similarly, if

1 R 1 þ R2 ¼ R R1 R2

then

R¼

R 1 R2 : R1 þ R2

Operations AF are sometimes called axioms of equality.

10.3

EQUIVALENT EQUATIONS

Equivalent equations are equations having the same solutions. Thus x  2 ¼ 0 and 2x ¼ 4 have the same solution x ¼ 2 and so are equivalent. However, x  2 ¼ 0 and x2  2x ¼ 0 are not equivalent, since x2  2x ¼ 0 has the additional solution x ¼ 0. The above operations used in transforming equations may not all yield equations equivalent to the original equations. The use of such operations may yield derived equations with either more or fewer solutions than the original equation. If the operations yield an equation with more solutions than the original, the extra solutions are called extraneous and the derived equation is said to be redundant with respect to the original equation. If the operations yield an equation with fewer solutions than the original, the derived equation is said to be defective with respect to the original equation. Operations A and B always yield equivalent equations. Operations C and E may give rise to redundant equations and extraneous solutions. Operations D and F may give rise to defective equations.

10.4

FORMULAS

A formula is an equation which expresses a general fact, rule, or principle. For example, in geometry the formula A ¼ pr 2 gives the area A of a circle in terms of its radius r. In physics the formula s ¼ 12 gt2 , where g is approximately 32.2 ft/s2, gives the relation between the distance s, in feet, which an object will fall freely from rest during a time t, in seconds. To solve a formula for one of the variables involved is to perform the same operations on both members of the formula until the desired variable appears on one side of the equation but not on the other side. Thus if F ¼ ma, we may divide by m to obtain a ¼ F/m and the formula is solved for a in terms of the other variables F and m. To check the work, substitute a ¼ F/m into the original equation to obtain F ¼ m(F/m), an identity.

CHAP. 10]

10.5

75

EQUATIONS IN GENERAL

POLYNOMIAL EQUATIONS

A monomial in a number of unknowns x, y, z, . . . has the form axp yq zr    where the exponents p, q, r, . . . are either positive integers or zero and the coefficient a is independent of the unknowns. The sum of the exponents p þ q þ r þ    is called the degree of the term in the unknowns x, y, z, . . . EXAMPLES 10.4.

3x2 z3 , 12 x4 , 6 are monomials. 3x2 z3 is of degree 2 in x, 3 in z, and 5 in x and z. 1 4 zero. 2 x is of fourth degree. 6 is of degree pffiffiffi 4y=x ¼ 4yx1 is not integral in x; 3x yz3 is not rational in y:

When reference is made to degree without specifying the unknowns considered, the degree in all unknowns is implied. A polynomial in various unknowns consists of terms each of which is rational and integral. The degree of such a polynomial is defined as the degree of the terms of highest degree. EXAMPLE 10.5.

3x3 y4 z þ xy2 z5  8x þ 3 is a polynomial of degree 3 in x, 4 in y, 5 in z, 7 in x and y, 7 in y and z, 6 in x and z, and 8 in x, y, and z.

A polynomial equation is a statement of equality between two polynomials. The degree of such an equation is the degree of the term of highest degree present in the equation. EXAMPLE 10.6.

xyz2 þ 3xz ¼ 2x3 y þ 3z2 is of degree 3 in x, 1 in y, 2 in z, 4 in x and y, 3 in y and z, 3 in x and z,

and 4 in x, y, and z.

It should be understood that like terms in the equation have been combined. Thus 4x3 y þ x2 z  xy2 ¼ 4x y þ z should be written x2 z  xy2 ¼ z. 3

An equation is called linear if it is of degree 1 and quadratic if it is of degree 2. Similarly the words cubic, quartic and quintic refer to equations of degree 3, 4, and 5 respectively. EXAMPLES 10.7.

2x þ 3y ¼ 7z is a linear equation in x, y, and z: x2  4xy þ 5y2 ¼ 10 is a quadratic equation in x and y: x3 þ 3x2  4x  6 ¼ 0 is a cubic equation in x:

A polynomial equation of degree n in the unknown x may be written a0 xn þ a1 xn1 þ a2 xn2 þ    þ an1 x þ an ¼ 0

a0 = 0

where a0 , a1 , . . . , an are given constants and n is a positive integer. As special cases we see that a0 x þ a1 ¼ 0 or ax þ b ¼ 0 a0 x2 þ a1 x þ a2 ¼ 0 or ax2 þ bx þ c ¼ 0 a0 x 3 þ a1 x 2 þ a2 x þ a3 ¼ 0 a0 x4 þ a1 x3 þ a2 x2 þ a3 x þ a4 ¼ 0

is is is is

of of of of

degree degree degree degree

1 2 3 4

(linear equation), (quadratic equation), (cubic equation), (quartic equation).

Solved Problems 10.1

Which of the following are conditional equations and which are identities? (a) 3x  (x þ 4) ¼ 2(x  2), 2x  4 ¼ 2x  4; identity.

76

EQUATIONS IN GENERAL

[CHAP. 10

(b) (x  1)(x þ 1) ¼ (x  1)2 , x2  1 ¼ x2  2x þ 1; conditional equation. (c) (y  3)2 þ 3(2y  3) ¼ y(y þ 1)  y, y2  6y þ 9 þ 6y  9 ¼ y2 þ y  y, y2 ¼ y2 ; identity. (d ) x þ 3y  5 ¼ 2(x þ 2y) þ 3, x þ 3y  5 ¼ 2x þ 4y þ 3; conditional equation.

10.2

Check each of the following equations for the indicated solution or solutions. (a)

x x þ ¼ 10; x ¼ 12: 2 3

12 12 þ ¼ 10, 6 þ 4 ¼ 10, and x ¼ 12 is a solution. 2 3

(b)

x2 þ 6x ¼ 3x  2; x ¼ 2, x ¼ 1: xþ2

22 þ 6(2) 16 ¼ 3(2)  2, ¼ 4, and x ¼ 2 is a solution. 2þ2 4

(1)2 þ 6(1) 5 ¼ 3(1)  2, ¼ 5, and x ¼ 1 is a solution. 1 þ 2 1 pffiffiffi (c) x2  xy þ y2 ¼ 19; x ¼ 2, y ¼ 3; x ¼ 4, y ¼ 2 þ 7; x ¼ 2, y ¼ 1: 2 2 x ¼ 2, y ¼ 3: p (2) pffiffi3ffi ¼ 19, 19 pffiffi¼ ffi 19, and x ¼ 2, yp¼ffiffiffi 3 is a solution. pffiffiffi ffiffiffi 2  (2)3 þ 7)ffiffiffiþ (2 þ 7)2 ¼ 19, 16  8  4 7 þ (4 þ 4 7 þ 7) ¼ x ¼ 4, y ¼ 2 þ 7: 4  4(2 þ p 19, 19 ¼ 19, and x ¼ 4, y ¼ 2 þ 7 is a solution. x ¼ 2, y ¼ 1: 22  2(1) þ (1)2 ¼ 19, 7 ¼ 19, and x ¼ 2, y ¼ 1 is not a solution.

10.3

Use the axioms of equality to solve each equation. (a) 2(x þ 3) ¼ 3(x  1), 2x þ 6 ¼ 3x  3: Transposing terms: 2x  3x ¼ 6  3,  x ¼ 9: Multiplying by1: x ¼ 9: Check: 2(9 þ 3) ¼ 3(9  1), 24 ¼ 24: x x (b) þ ¼ 1: 3 6 Multiplying by 6: 2x þ x ¼ 6, 3x ¼ 6: Dividing by 3: x ¼ 2: Check: 2=3 þ 2=6 ¼ 1, 1 ¼ 1: (c) 3y  2(y  1) ¼ 4(y þ 2), 3y  2y þ 2 ¼ 4y þ 8; y þ 2 ¼ 4y þ 8: Transposing: y  4y ¼ 8  2, 3y ¼ 6: Dividing by 3: y ¼ 6=(3) ¼ 2: Check: 3(2)  2(2  1) ¼ 4(2 þ 2), 0 ¼ 0: 2x  3 4x  5 (d ) ¼ : Multiplying by x  1, 2x  3 ¼ 4x  5 or x  1. x1 x1 Check: Substituting x ¼ 1 into the given equation, we find 1=0 ¼ 1=0. This is meaningless, since division by zero is an excluded operation, and the given equation has no solution. Note that (i)

2x  3 4x  5 ¼ x1 x1

and

(ii) 2x  3 ¼ 4x  5

are not equivalent equations. When (i) is multiplied by x  1 an extraneous solution x ¼ 1 is introduced, and equation (ii) is redundant with respect to equation (i). (e) x(x  3) ¼ 2(x  3). Dividing each member by x  3 gives a solution x ¼ 2. Now x  3 ¼ 0 or x ¼ 3 is also a solution of the given equation which was lost in the division. The required roots are x ¼ 2 and x ¼ 3. The equation x ¼ 2 is defective with respect to the given equation.

CHAP. 10]

77

EQUATIONS IN GENERAL

pffiffiffiffiffiffiffiffiffiffiffi x þ 2 ¼ 1. Squaring both sides, x þ 2 ¼ 1 or p x ffiffi¼ ffi 1. Check: Substituting x ¼ 1 into the given equation, 1 ¼ 1 or 1 ¼ 1 which is false. Thus x ¼ 1 is an extraneous solution. The given equation has no solution. pffiffiffiffiffiffiffiffiffiffiffiffiffi (g) 2x  4 ¼ 6. Squaring both sides,p2x ffiffiffiffiffi  4 ¼ 36 or x ¼ 20: p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Check: If x ¼ 20, 2(20)  4 ¼ 6 or 36 ¼ 6 which is true. Hence x ¼ 20 is a solution. In this case no extraneous root was introduced.

(f)

10.4

In each of the following formulas, solve for the indicated letter. (a) E ¼ IR, for R. (b)

Dividing both sides by I = 0, we have R ¼ E/I.

s ¼ v0 t þ 12 at2 , for a. Transposing, 12 at2 ¼ s 2

 v0 t: Multiplying by 2, at2 ¼ 2(s  v0 t):

Dividing by t = 0,

a¼ (c)

1 1 1 ¼ þ , for p: f p q

2(s  v0 t) : t2

Transposing, 1 1 1 qf ¼  ¼ : p f q fq

Taking reciprocals, p¼ pffiffiffiffiffiffiffi (d ) T ¼ 2p l=g, for g.

fq qf

(assuming q = f ):

Squaring both sides, T2 ¼

4p2 l : g

Multiplying by g, gT 2 ¼ 4p2 l. Dividing by T 2 = 0, g ¼ 4p2 l=T 2 .

10.5

In each of the following formulas, find the value of the indicated letter, given the values of the other letters. (a) F ¼ 95 C þ 32, F ¼ 68; find C:

68 ¼ 95 C þ 32, 36 ¼ 95 C; C ¼ 59 (36) ¼ 20:

Another method: 95C ¼ F  32, C ¼ 59(F  32) ¼ 59(68  32) ¼ 59(36) ¼ 20: (b)

1 1 1 1 1 1 52 1 ¼ þ , ¼ ¼ , R2 ¼ 10: ¼  6 15 R2 R2 6 15 30 10

1 1 1 ¼ þ , R ¼ 6, R1 ¼ 15; find R2 : R R1 R2 Another method: 1 1 1 R1  R ¼  ¼ , R2 R R1 RR1

4 (c) V ¼ pr 3 , 3

V ¼ 288p; find r:

Another method: 3V ¼ 4pr , 3

3V r ¼ , 4p 3

RR1 6(15) ¼ 10: ¼ R1  R 15  6 4 288p ¼ 216, 288p ¼ pr 3 , r3 ¼ 3 4p=3

R2 ¼

rffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffi 3 3ð288pÞ 3 3V 3 r¼ ¼ ¼ 216 ¼ 6. 4p 4p

r ¼ 6:

78

10.6

EQUATIONS IN GENERAL

[CHAP. 10

Determine the degree of each of the following equations in each of the indicated unknowns. (a) 2x2 þ xy  3 ¼ 0: x; y; x and y. Degree 2 in x, 1 in y, 2 in x and y. (b) 3xy2  4y2 z þ 5x  3y ¼ x4 þ 2: x; z; y and z; x, y, and z. Degree 4 in x, 1 in z, 3 in y and z, 4 in x, y, and z. 3 : x; x and z; x, y, and z. (c) x2 ¼ yþz As it stands it is not a polynomial equation. However, it can be transformed into one by multiplying by y þ z to obtain x2 (y þ z) ¼ 3 or x2 y þ x2 z ¼ 3. The derived equation is a polynomial equation of degree 2 in x, 3 in x and z, and 3 in x, y, and z. pffiffiffiffiffiffiffiffiffiffiffi (d ) x þ 3 ¼ x þ y: y; x and y. As given it is not a polynomial equation, but it can be transformed into one by squaring both sides. Thus we obtain x þ 3 ¼ x2 þ 2xy þ y2 , which is of degree 2 in y and 2 in x and y. It should be mentioned, however,p that theffi equations are not equivalent, since x2 þ 2xy þ y2 ¼ x þ 3 pffiffiffiffiffiffiffiffiffiffi ffi ffiffiffiffiffiffiffiffiffiffi includes both x þ 3 ¼ x þ y and  x þ 3 ¼ x þ y.

10.7

Find all values of x for which (a) x2 ¼ 81, (b) (x  1)2 ¼ 4. SOLUTION (a) There is nothing to indicate whether x is a positive or negative number,pso must ffiffiffiffiffiwe p pffiffiffiffiffiis possible. ffiffiffiffiffi assume either 2 ¼ 2 Taking the square root of both sides of the given equation,pwe obtain x 81 ¼ 9. Now ffiffiffiffiffi pffiffiffiffiffi x represents 2 2 a positive number (or zero) if x ffi is real. Hence we have x ¼ x if x is positive while x ¼ x if x is pffiffiffiffi negative. Thus when writing x2 we must consider that it is either x (if x . 0) or x (if x , 0). Therefore ffi pffiffiffiffi the equation x2 ¼ 9 may be written as either x ¼ 9 or x ¼ 9 (i.e. x ¼ 9). The two solutions may be written x ¼ +9. (b) (x  1)2 ¼ 4, +(x  1) ¼ 2 or (x  1) ¼ +2, and the two roots are x ¼ 3 and x ¼ 21.

10.8

Explain the fallacy in the following sequence of steps. (a) (b) (c) (d ) (e) (f) (g) (h)

Let x ¼ y: Multiply both sides by x: Subtract y2 from both sides: Write as: Divide by x 2 y: Replace x by its equal, y: Hence: Divide by y:

x¼y x2 ¼ xy x2  y2 ¼ xy  y2 (x 2 y)(x þ y) ¼ y(x 2 y) xþy¼y yþy¼y 2y ¼ y 2 ¼ 1.

SOLUTION There is nothing wrong in steps (a), (b), (c), (d). However, in (e) we divide by x  y, which from the original assumption is zero. Since division by zero is not defined, everything we do from (e) on is to be looked upon with disfavor.

10.9

Show that

pffiffiffi 2 is an irrational number, i.e., it cannot be the quotient of two integers.

SOLUTION pffiffiffi Assume that 2 ¼ p=q where p and q are integers having no common factor except +1, i.e., p/q is in lowest terms. Squaring, we have p2 =q2 ¼ 2 or p2 ¼ 2q2 . Since 2q2 is an even number, p2 is even and hence p is even (if p were odd, p2 would be odd); then p ¼ 2k, where k is an integer. Thus p2 ¼ 2q2 becomes (2k)2 ¼ 2q2 or q2 ¼ 2k2 ; hence

CHAP. 10]

79

EQUATIONS IN GENERAL

q2 is even and q is even. But if p and q are both even they wouldp have ffiffiffi a common factor 2, thus contradicting the assumption that they have no common factor except +1. Hence 2 is irrational.

Supplementary Problems 10.10

State which of the following are conditional equations and which are identities. ( f ) (x  3)(x2 þ 3x þ 9) ¼ x3  27

(a) 2x þ 3  (2  x) ¼ 4x  1 (b) (2y  1)2 þ (2y þ 1)2 ¼ (2y)2 þ 6

(g)

(c) 2{x þ 4  3(2x  1)} ¼ 3(4  3x) þ 2  x (d ) (x þ 2y)(x  2y)  (x  2y)2 þ 4y(2y  x) ¼ 0 (e) 10.11

9x2  4y2 ¼ 2x þ 3y 3x  2y

y2  4 ¼ 2y  1; y ¼ 3 y2

(e)

(g) x2  2y ¼ 3y2 ; x ¼ 4, y ¼ 2; x ¼ 1, y ¼ 1

(d ) x3  6x2 þ 11x  6 ¼ 0; 1, 2, 3

(h) (x þ y)2 þ (x  y)2 ¼ 2(x2 þ y2 ); any values of x, y

Use the axioms of equality to solve each equation. Check the solutions obtained. (a) 5(x  4) ¼ 2(x þ 1)  7 2y y  ¼2 3 6 1 3 (c) ¼ 8  y y

(b)

(d )

3x  2 x þ 2 ¼ x2 x2 pffiffiffiffiffiffiffiffiffiffiffiffiffi ( f ) 3x  2 ¼ 4

(e)

(g)

pffiffiffiffiffiffiffiffiffiffiffiffiffi 2x þ 1 þ 5 ¼ 0

(h)

p ffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2x  3 þ 1 ¼ 0

(i) (y þ 1)2 ¼ 16 ( j) (2x þ 1)2 þ (2x  1)2 ¼ 34

xþ1 x1 ¼ x1 x2

In each of the following formulas, solve for the indicated letter. P1 V1 P2 V2 ¼ ; T2 T1 T2 rffiffiffiffiffi 2s (b) t ¼ ;s g ffi 1pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2a2 þ 2b2  c2 ; c (c) m ¼ 2 (a)

10.14

1 1 1 þ ¼ ;x¼3 x 2x x  1

pffiffiffi ( f ) y3 þ y2  5y  5 ¼ 0; + 5, 1

(b) x2  3x ¼ 4; 1; 4 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (c) 3x  2  x þ 2 ¼ 4; 34, 2

10.13

(h) (x2  y2 )2 þ (2xy)2 ¼ (x2 þ y2 )2

Check each of the following equations for the indicated solution or solutions. (a)

10.12

x2 x2 þ ¼ x2 4 12

(d ) v2 ¼ v20 þ 2as; a rffiffiffiffi m (e) T ¼ 2p ;k k (f) S ¼

n ½2a þ (n  1)d ; d 2

In each formula find the value of the indicated letter given the values of the other letters. (a) v ¼ v0 þ at; find a if v ¼ 20, v0 ¼ 30, t ¼ 5. n (b) S ¼ (a þ d); find d if S ¼ 570, n ¼ 20, a ¼ 40. 2 1 1 1 ¼ þ ; find q if f ¼ 30, p ¼ 10. (c) f p q 1 (d ) Fs ¼ mv2 ; find v if F ¼ 100, s ¼ 5, m ¼ 2.5. 2 1 (e) f ¼ pffiffiffiffiffiffi; find C to four decimal places if f ¼ 1000, L ¼ 4  106 . 2p LC

80

10.15

EQUATIONS IN GENERAL

[CHAP. 10

Determine the degree of each equation in each of the indicated unknowns. (a) x3  3x þ 2 ¼ 0: x (b) x2 þ xy þ 3y4 ¼ 6: x; y; x and y (c) 2xy3  3x2 y2 þ 4xy ¼ 2x3 : x; y; x and y (d ) xy þ yz þ xz þ z2 x ¼ y4 : x; y; z; x and z; y and z; x, y, and z

10.16

Classify each equation according as it is (or can be transformed into) an equation which is linear, quadratic, cubic, quartic or quintic in all of the unknowns present. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) 2x4 þ 3x3  x  5 ¼ 0 (e) x2 þ y2  1 ¼ x þ y 2x þ y ¼4 x  3y

(b) x  2y ¼ 4

( f)

(c) 2x2 þ 3xy þ y2 ¼ 10

(g) 3y2  4y þ 2 ¼ 2(y  3)2

(d ) x2 y3  2xyz ¼ 4 þ y5

(h) (z þ 1)2 (z  2) ¼ 0

10.17

Is the equation

10.18

Prove that

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (x þ 4)2 ¼ x þ 4 an identity? Explain.

pffiffiffi 3 is irrational.

ANSWERS TO SUPPLEMENTARY PROBLEMS 10.10

(a) Conditional equation (b) Conditional equation (c) Identity

10.11

(a) (b) (c) (d )

10.12

(a) x ¼ 5 (b) y ¼ 4

y ¼ 3 is a solution. x ¼ 21 is a solution, x ¼ 24 is not. x ¼ 34 is a solution, x ¼ 2 is not. x ¼ 1, 2, 3 are all solutions.

10.13 (a) T2 ¼

(c) y ¼ 1/2 (d) x ¼ 3

P2 V2 T1 P1 V1

(b) s ¼ 12 gt2 10.14 (a) a ¼ 22 10.15 (a) 3 10.16

10.17

10.18

(d ) Identity (e) Conditional equation ( f ) Identity

(a) quartic (b) linear

x ¼ 3 ispaffiffiffi solution. y ¼ + 5, 1 are all solutions. x ¼ 4, y ¼ 2; x ¼ 1, y ¼ 1 are solutions. The equation is an identity; hence any values of x and y are solutions.

(e) no solution (f) x ¼ 6

(g) no solution (h) x ¼ 1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (c) c ¼ + 2a2 þ 2b2  4m2

4p2 m T2 2S  2an ( f) d ¼ n(n  1)

(d ) a ¼ (b) d ¼ 17

(b) 2, 4, 4

(e) ( f) (g) (h)

(g) Conditional equation (h) Identity

v2  v20 2s (c) q ¼ 215

(c) 3, 3, 4

(c) quadratic (d ) quintic

(i) y ¼ 3, 25 ( j) x ¼ +2

(e) k ¼

(d) v ¼ +20

(e) C ¼ 0.0063

(d ) 1, 4, 2, 3, 4, 4 (e) quadratic ( f ) linear

(g) quadratic (h) cubic

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (x þ 4)2 ¼ x þ 4 only if x þ 4  0; ðx þ 4)2 ¼ (x þ 4) if x þ 4  0. The given equation is not an identity. pffiffiffi Assume 3 ¼ p=q where p and q are integers having no common factor except +1. Squaring, we get p2 =q2 ¼ 3 or p2 ¼ 3q2 . So p2 is a multiple of 3 or p ¼ 3k where k is an integer (if p ¼ 3k þ 1 or p ¼ 3k þ 2, then p2 is not a multiple of 3). Thus p2 ¼ 3q2 becomes (3k)2 ¼ 3q2 and q2 ¼ 3k2 . Since q2 is a multiple of 3, q is a multiple of 3. But if both p and q are multiples of 3, then they have pffiffiffi a common factor of 3. This contradicts the assumption that they have no common factor except +1. Hence, 3 is irrational.

CHAPTER 11

Ratio, Proportion, and Variation 11.1

RATIO

The ratio of two numbers a and b, written a : b, is the fraction a/b provided b = 0. Thus a : b ¼ a/b, b = 0. If a ¼ b = 0, the ratio is 1 : 1 or 1/1 ¼ 1. EXAMPLES 11.1.

11.2

4 2 ¼ . 6 3 3y 5x 20x (3) 5x : ¼ ¼ 4 3y=4 3y

(1)

The ratio of 4 to 6 ¼ 4 : 6 ¼

(2)

2 4 2=3 5 : ¼ ¼ 3 5 4=5 6

PROPORTION

A proportion is an equality of two ratios. Thus a : b ¼ c : d, or a=b ¼ c=d, is a proportion in which a and d are called the extremes and b and c the means, while d is called the fourth proportional to a, b, and c. In the proportion a : b ¼ b : c, c is called the third proportional to a and b, and b is called a mean proportional between a and c. Proportions are equations and can be transformed using procedures for equations. Some of the transformed equations are used frequently and are called the laws of proportion. If a=b ¼ c=d, then (1)

ad ¼ bc

(3)

a b ¼ c d

(5)

ab cd ¼ b d

(2)

b d ¼ a c

(4)

aþb cþd ¼ b d

(6)

aþb cþd ¼ : ab cd

11.3

VARIATION

In reading scientific material, it is common to find such statements as “The pressure of an enclosed gas varies directly as the temperature.” This and similar statements have precise mathematical meanings and they 81

82

RATIO, PROPORTION, AND VARIATION

[CHAP. 11

represent a specific type of function called variation functions. The three general types of variation functions are direct, inverse, and joint. (1)

If x varies directly as y, then x ¼ ky or x/y ¼ k, where k is called the constant of proportionality or the constant of variation.

(2) (3)

If x varies directly as y2 , then x ¼ ky2 . If x varies inversely as y, then x ¼ k/y.

(4)

If x varies inversely as y2 , then x ¼ k=y2 :

(5) (6)

If x varies jointly as y and z, then x ¼ kyz. If x varies directly as y2 and inversely as z, then x ¼ ky2 =z:

The constant k may be determined if one set of values of the variables is known.

11.4

UNIT PRICE

When shopping we find that many items are sold in different sizes. To compare the prices, we must compute the price per unit of measure for each size of the item. EXAMPLES 11.2.

What is the unit price for each item?

(a) 3 ounce jar of olives costing 87¢ x¢ 1 oz ¼ 87¢ 3 oz

x¼

87 ¼ 29 3

x¼

132 ¼ 11 12

29¢ per oz

(b) 12 ounce box of cereal costing $1.32 x¢ 1 oz ¼ 132¢ 12 oz

EXAMPLES 11.3.

11¢ per oz

What is the unit price for each item to the nearest tenth of a cent?

(a) 6.5 ounce can of tuna costing $1.09 x¢ 1 oz ¼ 109¢ 6:5 oz

x¼

109 ¼ 16:8 6:5

16:8¢ per oz

x¼

195 ¼ 13:9 14

13:9¢ per oz

(b) 14 ounce can of salmon costing $1.95 x¢ 1 oz ¼ 195¢ 14 oz

11.5

BEST BUY

To determine the best buy, we compare the unit price for each size of the item and the size with the lowest unit price is the best buy. In doing this, we make two assumptions – a larger size will not result in any waste and the buyer can afford the total price for each of the sizes of the item. The unit price is usually rounded to the nearest tenth of a cent when finding the best buy.

CHAP. 11]

83

RATIO, PROPORTION, AND VARIATION

EXAMPLE 11.4. Which is the best buy on a bottle of vegetable oil when 1 gallon costs $5.99, 1 pint costs 89¢, and 24 ounces costs $1.29? a¢ 1 oz 599 ¼ a¼ ¼ 4:7 4:7¢ per oz 599¢ 128 oz 128 b¢ 1 oz ¼ 89¢ 16 oz c¢ 1 oz ¼ 129¢ 24 oz

89 ¼ 5:6 16 129 c¼ ¼ 5:4 24

b¼

5:6¢ per oz 5:4¢ per oz

The best buy is one gallon of vegetable oil for $5.99.

Solved Problems Ratio and Proportion 11.1

Express each of the following ratios as a simplified fraction. (a) 96 : 128 ¼

96 3 ¼ 128 4

(d) (xy2  x2 y) : (x  y)2 ¼ 11.2

(b)

2 3 2=3 8 : ¼ ¼ 3 4 3=4 9

(c) xy2 : x2 y ¼

xy2 y ¼ x2 y x

xy2  x2 y xy(y  x) xy ¼ ¼ 2 2 yx (x  y) (y  x)

Find the ratio of each of the following quantities. (a) 6 pounds to 12 ounces. It is customary to express the quantities in the same units. Then the ratio of 96 ounces to 12 ounces is 96 : 12 ¼ 8 : 1. (b) 3 quarts to 2 gallons. The required ratio is 3 quarts to 8 quarts or 3 : 8. (c) 3 square yards to 6 square feet. Since 1 square yard ¼ 9 square feet, the required ratio is 27 ft2 : 6 ft2 ¼ 9 : 2.

11.3

In each of the following proportions determine the value of x. 3x 2 1 ¼ and x ¼ : xþ1 1 3 x þ 3 3x  2 ¼ and x ¼ 2: (b) (x þ 3) : 10 ¼ (3x  2) : 8, 10 8 x  1 2x  4 ¼ , x2  5x ¼ 0, (c) (x  1) : (x þ 1) ¼ (2x  4) : (x þ 4), xþ1 xþ4 x(x  5) ¼ 0 and x ¼ 0, 5:

(a) (3  x) : (x þ 1) ¼ 2 : 1,

11.4

Find the fourth proportional to each of the following sets of numbers. In each case let x be the fourth proportional. 2 6 ¼ and x ¼ 9: (a) 2, 3, 6. Here 2 : 3 ¼ 6 : x, 3 x 25 (b) 4, 25, 10. Here 4 : 5 ¼ 10 : x and x ¼  : 2 2b (c) a2 , ab, 2: Here a2 : ab ¼ 2 : x, a2 x ¼ 2ab and x ¼ : a

84

11.5

11.6

RATIO, PROPORTION, AND VARIATION

[CHAP. 11

Find the third proportional to each of the following pairs of numbers. In each case let x be the third proportional. (a) 2, 3.

Here 2 : 3 ¼ 3 : x

8 (b) 22, : 3

Here 2 :

x ¼ 9=2:

and

8 8 ¼ :x 3 3

and

x¼

32 : 9

Find the mean proportional between 2 and 8. SOLUTION Let x be the required mean proportional. Then 2 : x ¼ x : 8, x2 ¼ 16 and x ¼ +4:

11.7

A line segment 30 inches long is divided into two parts whose lengths have the ratio 2 : 3. Find the lengths of the parts. SOLUTION Let the required lengths be x and 30 2 x. Then x 2 ¼ 30  x 3

11.8

and

x ¼ 12 in:,

30  x ¼ 18 in:

Two brothers are respectively 5 and 8 years old. In how many years (x) will the ratio of their ages be 3 : 4? SOLUTION In x years their respective ages will be 5 þ x and 8 þ x. Then (5 þ x) : (8 þ x) ¼ 3 : 4, 4(5 þ x) ¼ 3(8 þ x), and x ¼ 4.

11.9

Divide 253 into four parts proportional to 2, 5, 7, 9. SOLUTION Let the four parts be 2k, 5k, 7k, 9k. Then 2k þ 5k þ 7k þ 9k ¼ 253 and k ¼ 11. Thus the four parts are 22, 55, 77, 99.

11.10 If x : y : z ¼ 2 : 25 : 4 and x 2 3y þ z ¼ 63, find x, y, z. SOLUTION Let x ¼ 2k, y ¼ 25k, z ¼ 4k. Substitute these values in x  3y þ z ¼ 63 and obtain 2k  3(5k) þ 4k ¼ 63 or k ¼ 3. Hence x ¼ 2k ¼ 6, y ¼ 5k ¼ 15, z ¼ 4k ¼ 12.

Variation 11.11 For each of the following statements write an equation, employing k as the constant of proportionality. (a) The circumference C of a circle varies as its diameter d. Ans. C ¼ kd (b) The period of vibration T of pffia simple pendulum at a given place is proportional to the square root of its length l. Ans. T ¼ k l

CHAP. 11]

85

RATIO, PROPORTION, AND VARIATION

(c) The rate of emission of radiant energy E per unit area of a perfect radiator is proportional to the fourth power of its absolute temperature T. Ans. E ¼ kT 4 (d) The heat H in calories developed in a conductor of resistance R ohms when using a current of I amperes, varies jointly as the square of the current, the resistance of the conductor, and the time t during which the conductor draws the current. Ans. H ¼ kI 2 Rt (e) The intensity I of a sound wave varies jointly as the square of its frequency n, the square of its amplitude r, the speed of sound v, and the density d of the undisturbed medium. Ans. I ¼ kn2 r 2 vd ( f ) The force of attraction F between two masses m1 and m2 varies directly as the product of the masses and inversely as the square of the distance r between them. Ans. F ¼ km1 m2 =r 2 (g) At constant temperature, the volume V of a given mass of an ideal gas varies inversely as the pressure p to which it is subjected. Ans. V ¼ k/p (h) An unbalanced force F acting on a body produces in it an acceleration a which is directly proportional to the force and inversely proportional to the mass m of the body. Ans. a ¼ kF/m 11.12 The kinetic energy E of a body is proportional to its weight W and to the square of its velocity v. An 8 lb body moving at 4 ft/sec has 2 ft-lb of kinetic energy. Find the kinetic energy of a 3 ton (6000 lb) truck speeding at 60 mi/hr (88 ft/sec). SOLUTION To find k : E ¼ kWv 2

or

k¼

E 2 1 ¼ : ¼ Wv2 8(42 ) 64

Thus the kinetic energy of the truck is E ¼

Wv2 6000(88)2 ¼ ¼ 726 000 ft-lb. 64 64

11.13 The pressure p of a given mass of ideal gas varies inversely as the volume V and directly as the absolute temperature T. To what pressure must 100 cubic feet of helium at 1 atmosphere pressure and 2538 temperature be subjected to be compressed to 50 cubic feet when the temperature is 3138? SOLUTION pV 1(100) 100 ¼ ¼ : T 253 253   100 T 100 313 Thus the required pressure is p ¼ ¼ ¼ 2:47 atmospheres. 253 V 253 50 To find k : p ¼ k

T V

or

k¼

Another method: Let the subscripts 1 and 2 refer to the initial and final conditions of the gas, respectively. Then

k¼

p1 V1 p2 V2 ¼ , T1 T2

p1 V1 p2 V2 ¼ , T1 T2

1(100) p2 (50) ¼ 253 313

and

p2 ¼ 2:47 atm:

11.14 If 8 men take 12 days to assemble 16 machines, how many days will it take 15 men to assemble 50 machines? SOLUTION The number of days (x) varies directly as the number of machines ( y) and inversely as the number of men (z). Then

x¼

ky z

Hence the required number of days is x ¼

where

k¼

xz 12(8) ¼ ¼ 6: y 16

6y 6(50) ¼ ¼ 20 days: z 15

86

RATIO, PROPORTION, AND VARIATION

[CHAP. 11

Unit Price and Best Buy

11.15 What is the unit price for 12 oranges costing 99¢? SOLUTION x¢ 1 orange ¼ 99¢ 12 oranges

x¼

99 ¼ 8:25 12

8:25¢ per orange

11.16 What is the unit price for trash bags when 20 bags cost $2.50? SOLUTION x¢ 1 bag ¼ 250¢ 20 bags

x¼

250 ¼ 12:5 20

12:5¢ per bag

11.17 Which is the best buy when 7 cans of soup cost $2.25 and 3 cans of soup cost 95¢? SOLUTION a¢ 1 can ¼ 225¢ 7 cans b¢ 1 can ¼ 95¢ 3 cans

225 ¼ 32:1 7 95 b¼ ¼ 31:7 3 a¼

32:1¢ per can 31:7¢ per can

The best buy is 3 cans of soup costing 95¢.

11.18 Which is the best buy when a 3 ounce package of cream cheese costs 43¢ and an 8 ounce package of cream cheese costs 87¢? SOLUTION a¢ 1 oz ¼ 43¢ 3 oz b¢ 1 oz ¼ 87¢ 8 oz

43 ¼ 14:3 3 87 b¼ ¼ 10:9 8 a¼

14:3¢ per oz 10:9¢ per oz

The best buy is the 8 ounce package of cream cheese costing 87¢.

Supplementary Problems 11.19

Express each ratio as a simplified fraction. (a) 40 : 64

11.20

(b) 4/5 : 8/3

(d ) (a2 b þ ab2 ) : (a2 b3 þ a3 b2 )

Find the ratio of the following quantities. (a) 20 yards to 40 feet (b) 8 pints to 5 quarts

11.21

(c) x2 y3 : 3xy4

(c) 2 square feet to 96 square inches (d ) 6 gallons to 30 pints

In each proportion determine the value of x. (a) (x þ 3) : (x  2) ¼ 3 : 2 (b) (x þ 4) : 1 ¼ (2  x) : 2

(c) (x þ 1) : 4 ¼ (x þ 6) : 2x (d ) (2x þ 1) : (x þ 1) ¼ 5x : (x þ 4)

CHAP. 11]

11.22

87

RATIO, PROPORTION, AND VARIATION

Find the fourth proportional to each set of numbers. (a) 3, 4, 12

(b) 22, 5, 6

(c) a, b, c

(d ) m þ 2, m 2 2, 3

11.23

Find the third proportional to each pair of numbers. pffiffiffiffiffi (a) 3, 5 (b) 22, 4 (c) a, b (d ) ab, ab

11.24

Find the mean proportional between each pair of numbers. pffiffiffi pffiffiffi (a) 3, 27 (b) 24, 28 (c) 3 2 and 6 2 (d ) m þ 2 and

mþ1

11.25

If (x þ y) : (x 2 y) ¼ 5 : 2, find x : y.

11.26

Two numbers have the ratio 3 : 4. If 4 is added to each of the numbers the resulting ratio is 4 : 5. Find the numbers.

11.27

A line segment of length 120 inches is divided into three parts whose lengths are proportional to 3, 4, 5. Find the lengths of the parts.

11.28

If x : y : z ¼ 4 : 23 : 2 and 2x þ 4y 2 3z ¼ 20, find x, y, z.

11.29

(a) If x varies directly as y and if x ¼ 8 when y ¼ 5, find y when x ¼ 20. (b) If x varies directly as y2 and if x ¼ 4 when y ¼ 3, find x when y ¼ 6. (c) If x varies inversely as y and if x ¼ 8 when y ¼ 3, find y when x ¼ 2.

11.30

The distance covered by an object falling freely from rest varies directly as the square of the time of falling. If an object falls 144 ft in 3 sec, how far will it fall in 10 sec?

11.31

The force of wind on a sail varies jointly as the area of the sail and the square of the wind velocity. On a square foot of sail the force is 1 lb when the wind velocity is 15 mi/hr. Find the force of a 45 mi/hr wind on a sail of area 20 square yards.

11.32

If 2 men can plow 6 acres of land in 4 hours, how many men are needed to plow 18 acres in 8 hours?

11.33

What is the unit price to the nearest tenth of a cent for each item? (a) (b) (c) (d ) (e) ( f)

11.34

1.36 liter can of fruit punch costing $1.09 283 gram jar of jelly costing 79¢ 10.4 ounce jar of face cream costing $3.73 1 dozen cans of peas costing $4.20 25 pounds of grass seed costing $27.75 3 doughnuts costing 49¢

Which is the best buy? (a) 100 aspirin tablets for $1.75 or 200 aspirin tablets for $2.69 (b) a 6 ounce jar of peanut butter for 85¢ or a 12 ounce jar of peanut butter for $1.59 (c) a 14 ounce bottle of mouthwash for $1.15 or a 20 ounce bottle of mouthwash for $1.69 (d ) a 9 ounce jar of mustard for 35¢ or a 24 ounce jar of mustard for 89¢ (e) a 454 gram box of crackers for $1.05 or a 340 gram box of crackers for 93¢ ( f ) a 0.94 liter bottle of fabric softener for 99¢ or a 2.76 liter bottle of fabric softener for $2.65

ANSWERS TO SUPPLEMENTARY PROBLEMS 11.19

(a) 5/8 (b) 3/10 (c) x/3y

(d) 1/ab

11.20

(a) 3 : 2 (b) 4 : 5 (c) 3 : 1 (d) 8 : 5

88

RATIO, PROPORTION, AND VARIATION

11.21

(a) 12 (b) 22

11.22

(a) 16 (b) 215

11.23 11.24

(a) 25/3 (b) 28 (c) b 2/a (d ) 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (a) +9 (b) +4 2 (c) +6 (d ) + m2 þ 3m þ 2

11.25

7/3

11.26

12, 16

11.27

30, 40, 50 in.

11.28

28, 6, 24

11.29

(a) 12 12 (b) 16 (c) 12

11.30

1600 ft

11.31

1620 lb

11.32

3 men

11.33

(a) 80.1¢ per liter (b) 0.3¢ per gram

11.34

(a) 200 aspirins for $2.69 (b) 12 ounce jar for $1.59

[CHAP. 11

(c) 4, 23 (d) 2, 22/3 (c) bc/a (d ) 3(m  2)=(m þ 2)

(c) 35.9¢ per ounce (d ) 35¢ per can

(e) 111¢ per pound ( f ) 16.3¢ per doughnut

(c) 14 ounce bottle for $1.15 (d ) 24 ounce jar for 89¢

(e) 454 gram box for $1.05 ( f ) 2.76 liter bottle for $2.65

CHAPTER 12

Functions and Graphs 12.1

VARIABLES

A variable is a symbol which may assume any one of a set of values during a discussion. A constant is a symbol which stands for only one particular value during the discussion. Letters at the end of the alphabet, such as x, y, z, u, v, and w, are usually employed to represent variables, and letters at the beginning of the alphabet, such as a, b, and c, are used as constants. 12.2

RELATIONS

A relation is a set of ordered pairs. The relation may be specified by an equation, a rule, or a table. The set of the first components of the ordered pairs is called the domain of the relation. The set of the second components is called the range of the relation. In this chapter, we shall consider only relations that have sets of real numbers for their domain and range. EXAMPLE 12.1.

What is the domain and range of the relation {(1, 3), (2, 6), (3, 9), (4, 12)}? domain ¼ {1, 2, 3, 4}

12.3

range ¼ {3, 6, 9, 12}

FUNCTIONS

A function is a relation such that each element in the domain is paired with exactly one element in the range. EXAMPLES 12.2.

Which relations are functions?

(a) f(1, 2), (2, 3), (3, 4), (4, 5)g function – each first element is paired with exactly one second element (b) f(1, 2), (1, 3), (2, 8), (3, 9)g not a function – 1 is paired with 2 and with 3 (c) f(1, 3), (2, 3), (4, 3), (9, 3)g function – each first element is paired with exactly one second element

Often functions and relations are stated as equations. When the domain is not stated, we determine the largest subset of the real numbers for which the equation is defined, and that is the domain. Once the domain has been determined, we determine the range by finding the value of the equation for each value of the domain. The variable associated with the domain is called an independent variable and the variable 89

90

FUNCTIONS AND GRAPHS

[CHAP. 12

associated with the range is called the dependent variable. In equations with the variables x and y, we generally assume that x is the independent variable and that y is the dependent variable. EXAMPLE 12.3. What is the domain and range of y ¼ x2 þ 2? The domain is the set of all real numbers since the square of each real number is a real number and that a real number plus 2 is still a real number. Domain ¼ {all real numbers} The range is the set of all real numbers greater than or equal to 2 since the square of a real number is at least zero and when you add 2 to each value we have the real numbers that are at least 2. Range ¼ {all real numbers 2} EXAMPLE 12.4. What is the domain and range of y ¼ 1/(x 2 3)? The equation is not defined when x ¼ 3, so the domain is the set of all real numbers not equal to 3. Domain ¼ {real numbers = 3} A fraction can be zero only when the numerator can be zero. Since the numerator of this fraction is always 1, the fraction can never equal zero. Thus, the range is the set of all real numbers not equal to 0. Range ¼ {real numbers = 0}.

12.4

FUNCTION NOTATION

The notation y ¼ f (x), read “y equals f of x,” is used to designate that y is a function of x. With this notation f (a) represents the value of the dependent variable y when x ¼ a (provided that there is a value). Thus y ¼ x2  5x þ 2 may be written f (x) ¼ x2  5x þ 2. Then f (2), i.e., the value of f (x) or y when x ¼ 2, is f (2) ¼ 22  5(2) þ 2 ¼ 4. Similarly, f (1) ¼ (1)2  5(1) þ 2 ¼ 8. Any letter may be used in the function notation; thus g(x), h(x), F(x), etc., may represent functions of x.

12.5

RECTANGULAR COORDINATE SYSTEM

A rectangular coordinate system is used to give a picture of the relationship between two variables. Consider two mutually perpendicular lines X 0 X and Y 0 Y intersecting in the point O, as shown in Fig. 12-1.

Fig. 12-1

The line X 0 X, called the x axis, is usually horizontal. The line Y 0 Y, called the y axis, is usually vertical. The point O is called the origin.

CHAP. 12]

91

FUNCTIONS AND GRAPHS

Using a convenient unit of length, lay off points on the x axis at successive units to the right and left of the origin O, labeling those points to the right 1, 2, 3, 4, . . . and those to the left 21,22,23,24, . . . . Here we have arbitrarily chosen OX to be the positive direction; this is customary but not necessary. Do the same on the y axis, choosing OY as the positive direction. It is customary (but not necessary) to use the same unit of length on both axes. The x and y axes divide the plane into 4 parts known as quadrants, which are labeled I, II, III, IV as in Fig. 12-1. Given a point P in this xy plane, drop perpendiculars from P to the x and y axes. The values of x and y at the points where these perpendiculars meet the x and y axes determine respectively the x coordinate (or abscissa) of the point and the y coordinate (or ordinate) of the point P. These coordinates are indicated by the symbol (x, y). Conversely, given the coordinates of a point, we may locate or plot the point in the xy plane. For example, the point P in Fig. 12-1 has coordinates (3, 2); the point having coordinates (22, 23) is Q. The graph of a function y ¼ f (x) is the set of all points (x, y) satisfied by the equation y ¼ f (x). 12.6

FUNCTION OF TWO VARIABLES

The variable z is said to be a function of the variables x and y if there exists a relation such that to each pair of values of x and y there corresponds one or more values of z. Here x and y are independent variables and z is the dependent variable. The function notation used in this case is z ¼ f (x, y): read “z equals f of x and y.” Then f (a, b) denotes the value of z when x ¼ a and y ¼ b, provided the function is defined for these values. Thus if f (x, y) ¼ x3 þ xy2  2x, then f (2, 3) ¼ 23 þ 2  32  2  3 ¼ 20: In like manner we may define functions of more than two independent variables. 12.7

SYMMETRY

When the left half of a graph is a mirror image of the right half, we say the graph is symmetric with respect to the y axis (see Fig. 12-2). This symmetry occurs because for any x value, both x and 2x result in the same y value, that is f (x) ¼ f (x). The equation may or may not be a function for y in terms of x. Some graphs have a bottom half that is the mirror image of the top half, and we say these graphs are symmetric with respect to the x axis. Symmetry with respect to the x axis results when for each y, both y and 2y result in the same x value (see Fig. 12-3). In these cases, you do not have a function for y in terms of x. If substituting for 2x for x and 2y for y in an equation yields an equivalent equation, we say the graph is symmetric with respect to the origin (see Fig. 12-4). These equations represent relations that are not always functions.

Fig. 12-2

Fig. 12-3

Fig. 12-4

92

FUNCTIONS AND GRAPHS

[CHAP. 12

Symmetry can be used to make sketching the graphs of relations and functions easier. Once the type of symmetry, if any, and the shape of half of the graph have been determined, the other half of the graph can be drawn by using this symmetry. Most graphs are not symmetric with respect to the y axis, x axis, or the origin. However, many frequently used graphs do display one of these symmetries and using that symmetry in graphing the relation simplifies the graphing process. EXAMPLE 12.5. Test the relation y ¼ 1/x for symmetry. Substituting 2x for x, we get y ¼ 1=x, so the graph is not symmetric with respect to the y axis. Substituting 2y for y, we get y ¼ 1=x, so the graph is not symmetric with respect to the x axis. Substituting 2x for x and 2y for y, we get y ¼ 1=x which is equivalent to y ¼ 1=x, so the graph is symmetric with respect to the origin.

12.8

SHIFTS

The graph of y ¼ f (x) is shifted upward by adding a positive constant to each y value in the graph. It is shifted downward by adding a negative constant to each y value in the graph of y ¼ f (x): Thus, the graph of y ¼ f (x) þ b differs from the graph of y ¼ f (x) by a vertical shift of jbj units. The shift is up if b . 0 and the shift is down if b , 0. EXAMPLES 12.6. How do the graphs of y ¼ x2 þ 2 and y ¼ x2  3 differ from the graph of y ¼ x2 ? The graph of y ¼ x2 is shifted up 2 units to yield the graph of y ¼ x2 þ 2 (see Figs. 12-5(a) and (b)). The graph of y ¼ x2 is shifted 3 units down to yield the graph of y ¼ x2  3 (see Figs. 12-5(a) and (c)).

Fig. 12-5

The graph of y ¼ f (x) is shifted to the right when a positive number is subtracted from each x value. It is shifted to the left if a negative number is subtracted from each x value. Thus, the graph of y ¼ f (x  a) differs from the graph of y ¼ f ðxÞ by a horizontal shift of j a j units. The shift is to the right if a . 0 and the shift is to the left if a , 0: EXAMPLES 12.7. How do the graphs of y ¼ (x þ 1)2 and y ¼ (x  2)2 differ from the graph of y ¼ x2 ? The graph of y ¼ x2 is shifted 1 unit to the left to yield the graph of y ¼ (x þ 1)2 since x þ 1 ¼ x  (1) (see Figs. 12-6(a) and (b)).

CHAP. 12]

FUNCTIONS AND GRAPHS

93

The graph of y ¼ x2 is shifted 2 units to the right to yield the graph of y ¼ (x  2)2 (see Figs. 12-6(a) and (c)).

Fig. 12-6

12.9

SCALING

If each y value is multiplied by a positive number greater than 1, the rate of change in y is increased from the rate of change in y values for y ¼ f (x): However, if each y value is multiplied by a positive number between 0 and 1, the rate of change in y values is decreased from the rate of change in y values for y ¼ f (x). Thus, the graph of y ¼ cf (x), where c is a positive number, differs from the graph of y ¼ f (x) by the rate of increase in y. If c . 1 the rate of change in y is increased and if 0 , c , 1 the rate of change in y is decreased. The graph of y ¼ f (x) is reflected across the x axis when each y value is multiplied by a negative number. So the graph of y ¼ cf (x), where c , 0, is the reflection of y ¼j c j f (x) across the x axis. EXAMPLES 12.8. How do the graphs of y ¼  j x j, y ¼ 3 j x j, and y ¼ 1=2 j x j differ from the graph of y ¼ j x j? The graph of y ¼ j x j is reflected across the x axis to yield y ¼  j x j (see Figs. 12-7(a) and (b)). The graph of y ¼ j x j has the y value multiplied by 3 for each x value to yield the graph of y ¼ 3 j x j (see Figs. 12-7(a) and (c)). The graph of y ¼ j x j has the y value multiplied by 1=2 for each x value to yield the graph of y ¼ 1=2 j x j (see Figs. 12-7(a) and (d)).

12.10

USING A GRAPHING CALCULATOR

In discussing graphing calculators the information given will be general, but a Texas Instruments TI-84 graphing calculator was used to verify the general procedures. Most graphing calculators operate in a somewhat similar manner to one another, but you need to use the instruction manual for your calculator to see how to do these operations for that particular make and model of graphing calculator. A graphing calculator allows you to graph functions easily. The key to graphing is setting the graphing window appropriately. To do this, you need to use the domain of the function to set the maximum and minimum x values and the range to set the maximum and minimum y values. When the domain or range is a large interval, it may be necessary to use the scale for x or y to make the graph smaller, increasing the size of the units along either or both axes. Occasionally, it may be necessary to view the graph in parts of its domain or range to see how the graph actually looks. To compare the graphs of y ¼ x2 , y ¼ x2 þ 2, and y ¼ x2  3 on a graphing calculator, you enter each function in the y ¼ menu. Let y1 ¼ x2 , y2 ¼ x2 þ 2, and y3 ¼ x2  3. Turn off the functions y2 and y3 and set the graphing window at the standard setting. When you press the graph key, you will see a graph as shown in Fig. 12-5(a). Turn off the function y1 and turn on the function y2 , then press the graph key. The graph displayed will be Fig. 12-5(b). Now turn off function y2 and turn on function y3 . Press the graph key and you will see

94

FUNCTIONS AND GRAPHS

[CHAP. 12

Fig. 12-5(c). When you turn on functions y1 , y2 , and y3 and press the graph key and you will see all three functions graphed on the same set of axes (see Fig. 12-8). The graph of y2 ¼ x2 þ 2 is 2 units above the graph of y1 ¼ x2 , while the graph of y3 ¼ x2  3 is 3 units below the graph of y1 .

Fig. 12-7

In a similar fashion, you can compare the graph of y1 ¼ f (x) and y2 ¼ f (x) þ b for any function f (x). Notice that when b . 0, the graph of y2 is b units above the graph of y1 : When b , 0, the graph of y2 is j b j units below the graph of y1 : Consider the graphs of y ¼ x2 , y ¼ (x þ 1)2 , and y ¼ (x  2)2 . To compare these graphs using a calculator, we need to set y1 ¼ x2 , y2 ¼ (x þ 1)2 , and y3 ¼ (x  2)2 . Using the standard window and graphing all three functions at once, we see that y2 ¼ (x þ 1)2 is 1 unit to the left of the graph of y1 ¼ x2 . Also, the graph of y3 ¼ (x  2)2 is 2 units to the right of the graph of y1 (see Fig. 12-9). In general, to compare the graphs of y1 ¼ f (x) and y2 ¼ f (x  a) for all functions f (x), we note that the graph of y2 ¼ f (x  a) is a units to the right of y1 ¼ f ðxÞ when a . 0. When a , 0, the graph of y2 is j a j units to the left of y1 .

CHAP. 12]

FUNCTIONS AND GRAPHS

95

Fig. 12-8

Fig. 12-9 EXAMPLE 12.9. Graph x2 þ y2 ¼ 9. pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi To graph x2 þ ffiy2 ¼ 9 on a calculator, we first solve the equation for y and get y ¼ + 9  x2 . We let y1 ¼ þ 9  x2 pffiffiffiffiffiffiffiffiffiffiffiffi and y2 ¼  9  x2 and graph them on the same set of axes. If we use the standard window, we get a distorted view of this graph because the scale on the y axis is not equal to the scale on the x axis. By multiplying by a factor of 0.67 (for the TI-84), we can adjust the y interval and get a more accurate view of the graph. Thus using the domain ½10, 10 and a range of ½6:7, 6:7, we get the graph of a circle (see Fig. 12-10).

Fig. 12-10

96

FUNCTIONS AND GRAPHS

[CHAP. 12

Solved Problems 12.1

Express the area A of a square as a function of its (a) side x, (b) perimeter P, and (c) diagonal D (see Fig. 12-11). SOLUTION (a) A ¼ x2

Fig. 12-11  2 P P P2 : Then A ¼ x2 ¼ or A ¼ : 16 4 4  2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi D D D2 2 (c) D ¼ x2 þ x2 ¼ x 2 or x ¼ pffiffiffi : Then A ¼ x ¼ pffiffiffi or A ¼ : 2 2 2

(b) P ¼ 4x or x ¼

12.2

Express the (a) area A, (b) perimeter P, and (c) diagonal D of a rectangle as a function of its sides x and y. Refer to Fig. 12-12. SOLUTION (a) A ¼ xy,

(b) P ¼ 2x þ 2y,

(c) D ¼

Fig. 12-12

12.3

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y2

Fig. 12-13

Express the (a) altitude h and (b) area A of an equilateral triangle as a function of its side s. Refer to Fig. 12-13. SOLUTION sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi  2ffi rffiffiffiffiffiffiffiffi 1 3 2 s 3 s ¼ s ¼ (a) h ¼ s2  2 4 2

pffiffiffi  pffiffiffi 1 1 s 3 s2 3 (b) A ¼ hs ¼ s¼ 4 2 2 2

CHAP. 12]

12.4

97

FUNCTIONS AND GRAPHS

The surface area S and volume V of a sphere of radius r are given by S ¼ 4pr 2 and V ¼ 43 pr 3 . Express (a) r as a function of S and also as a function of V, (b) V as a function of S, and (c) S as a function of V. SOLUTION (a) From S ¼ 4pr2 obtain r¼

rffiffiffiffiffiffi rffiffiffiffi S 1 S ¼  4p 2 p

From V ¼ 43 pr 3 obtain r¼

rffiffiffiffiffiffi 3 3V  4p

r¼

rffiffiffiffi 1 S 2 p

(b) Set

in V ¼ 43 pr 3 and obtain

rffiffiffiffi!3 rffiffiffiffi 4 1 S S S V¼ p ¼  3 2 p 6 p

(c) Set rffiffiffiffiffiffi 3 3V r¼ 4p in S ¼ 4pr 2 and obtain

12.5

sffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi 2 4p p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3V 3 9V 3 S ¼ 4p ¼ 36pV 2  ¼ 4p  16p2 4p 4p

Given y ¼ 3x2  4x þ 1, find the values of y corresponding to x ¼ 22, 21, 0, 1, 2. SOLUTION For x ¼ 2, y ¼ 3(2)2  4(2) þ 1 ¼ 21; for x ¼ 2l, y ¼ 3(1)2  4(1) þ 1 ¼ 8; for x ¼ 0, y ¼ 3(0)2  4(0) þ 1 ¼ 1; for x ¼ 1, y ¼ 3(1)2  4(1) þ 1 ¼ 0; for x ¼ 2; y ¼ 3ð2Þ2  4ð2Þ þ 1 ¼ 5: These values of x and y are conveniently listed in the following table.

12.6

x

22

21

0

1

2

y

21

8

1

0

5

Extend the table of values in Problem 12.5 by finding the values of y corresponding to x ¼ 3=2, 1=2, 1=2, 3=2: SOLUTION For x ¼ 3=2, y ¼ 3(3=2)2  4(3=2) þ 1 ¼ 13 34; etc. The following table of values summarizes the results. x y

22

 32

21

13 34

21

 12

8

3 34

0

1 2

1

 14

1

3 2

2

0

1 34

5

98

12.7

FUNCTIONS AND GRAPHS

State the domain and range for each relation. pffiffiffiffiffiffiffiffiffiffiffi (b) y ¼ x3 þ 1 (c) y ¼ x þ 2 (a) y ¼ 3  x2

[CHAP. 12

(d ) y ¼

p ffiffiffi 3 x

SOLUTION (a) Domain ¼ {all real numbers} Since every real number can be squared, 3  x2 is defined for all real numbers. Range ¼ {all real numbers  3} Since x2 is non-negative for all real numbers, 3  x2 does not exceed 3. (b) Domain ¼ {all real numbers} Since every real number can be cubed, x3 þ 1 is defined for all real numbers. Range ¼ {all real numbers} Since x3 yields all real numbers, x3 þ 1 also yields all real numbers. (c) Domain ¼ {all real numbers  2} Since the square root yields real numbers only for non-negative real numbers, x must be at least 22. Range ¼ {all real numbers  0} Since we want the principal square root, the values will be non-negative numbers. (d ) Domain ¼ {all real numbers} Since the cube root yields a real number for all real numbers, x can be any real number. Range ¼ {all real numbers} Since any real number can be the cube root of a real number, we get all real numbers.

12.8

In which of these equations is y a function of x? pffiffiffiffiffi (a) y ¼ 3x3 (c) xy ¼ 1 (e) y ¼ 4x (d ) y ¼ 2x þ 5 ( f ) y3 ¼ 8x (b) y2 ¼ x

SOLUTION (a) (b) (c) (d ) (e) (f )

12.9

Function Not a function Function Function Function Function

For each value of x, 3x3 yields exactly one value. For x ¼ 4, y can be 2 or 22. y ¼ 1/x. For every non-zero real number 1/x yields exactly one value. For each value of x, 2x þp 5 ffiffiffiffiffi yields exactly one value. For each value of x  0, 4x yields the principal square root. For each real number, 8x is a real number and every real number has exactly one real cube root.

If f (x) ¼ x3  5x  2, find f (2), f (3=2), f (1), f (0), f (1), f (2):

SOLUTION f (2) ¼ (2)3  5(2)  2 ¼ 0 f (3=2) ¼ (3=2)3  5(3=2)  2 ¼ 17=8 f (1) ¼ (1)3  5(1) 2 ¼ 2

f (0) ¼ 03  5(0)  2 ¼ 2 f (1) ¼ 13  5(1)  2 ¼ 6 f (2) ¼ 23  5(2)  2 ¼ 4

We may arrange these values in a table.

12.10 If F(t) ¼

x

22

3=2

21

0

1

2

f (x)

0

17=8

2

22

26

24

t3 þ 2t , find F(2), F(x), F(x): t1

CHAP. 12]

FUNCTIONS AND GRAPHS

99

SOLUTION F(2) ¼ F(x) ¼ F(x) ¼

(2)3 þ 2(2) 8  4 ¼ ¼4 2  1 3 x3 þ 2x x1 (x)3 þ 2(x) x3  2x x3 þ 2x ¼ ¼ x  1 x  1 xþ1

12.11 Given R(x) ¼ (3x  1)=(4x þ 2), find   x1 R(x þ h)  R(x) , (a) R , (b) xþ2 h

(c) R½R(x):

SOLUTION

    x1 2x  5   3 1 x1 2x  5 x þ 2 x þ 2 ¼ ¼  ¼  (a) R x1 6x xþ2 6x þ2 4 xþ2 xþ2 (b)

  R(x þ h)  R(x) 1 1 3(x þ h)  1 3x  1 ¼ {R(x þ h)  R(x)} ¼  h h h 4(x þ h) þ 2 4x þ 2 ¼

  1 ½3(x þ h)  1½4x þ 2  ½3x  1½4(x þ h) þ 2 5 ¼ h ½4(x þ h) þ 2½4x þ 2 2(2x þ 2h þ 1)(2x þ 1)

  3x  1   3 1 3x  1 5x  5 x  1 4x þ 2  ¼  ¼ ¼ (c) R½R(x) ¼ R 3x  1 4x þ 2 20x 4x þ2 4 4x þ 2

12.12 If F(x, y) ¼ x3  3xy þ y2 , find (a) F(2, 3),

(b) F(3, 0),

(c)

F(x, y þ k)  F(x, y) : k

SOLUTION (a) F(2, 3) ¼ 23  3(2)(3) þ 32 ¼ 1 (b) F(3, 0) ¼ (3)3  3(3)(0) þ 02 ¼ 27 F(x, y þ k)  F(x, y) x3  3x(y þ k) þ (y þ k)2  ½x3  3xy þ y2  (c) ¼ ¼ 3x þ 2y þ k k k

12.13 Plot the following points on a rectangular coordinate system: (2, 1), (4, 3), (22, 4), (24, 2), (24, 22), pffiffiffi (25/2, 29/2), (4, 23), (2,  2): SOLUTION See Fig. 12-14.

12.14 Given y ¼ 2x  1, obtain the values of y corresponding to x ¼ 23, 22, 21, 0, 1, 2, 3 and plot the points (x, y) thus obtained.

100

FUNCTIONS AND GRAPHS

Fig. 12-14

[CHAP. 12

Fig. 12-16

Fig. 12-15

SOLUTION The following table lists the values of y corresponding to the given values of x. x

23

22

21

0

1

2

3

y

27

25

23

21

1

3

5

The points (3, 7), (  2, 5), (1, 3), (0, 1), (1, 1), (2, 3), (3, 5) are plotted, as shown in Fig. 12-15. Note that all points satisfying y ¼ 2x  1 lie on a straight line. In general the graph of y ¼ ax þ b, where a and b are constants, is a straight line; hence y ¼ ax þ b or f (x) ¼ ax þ b is called a linear function. Since two points determine a straight line, only two points need be plotted and the line drawn connecting them.

12.15 Obtain the graph of the function defined by y ¼ x2  2x  8 or f (x) ¼ x2  2x  8.

SOLUTION The following table gives the values of y or f(x) for various values of x. x

24

23

22

21

0

1

2

3

4

5

6

y or f (x)

16

7

0

25

28

29

28

25

0

7

16

Thus the following points lie on the graph: (4, 16), (3, 7), (2, 0), (1, 5), etc. In plotting these points it is convenient to use different scales on the x and y axes, as shown in Fig. 12-16. The points marked  were added to those already obtained in order to get a more accurate picture. The curve thus obtained is called a parabola. The lowest point P, called a minimum point, is the vertex of the parabola.

12.16 Graph the function defined by y ¼ 3  2x  x2 .

SOLUTION x

25

24

23

22

21

0

1

2

3

4

y

212

25

0

3

4

3

0

25

212

221

CHAP. 12]

101

FUNCTIONS AND GRAPHS

The curve obtained is a parabola, as shown in Fig. 12-17. The point Q(1, 4), the vertex of the parabola, is a maximum point. In general, y ¼ ax2 þ bx þ c represents a parabola whose vertex is either a maximum or minimum point depending on whether a is 2 or þ, respectively. The function f (x) ¼ ax2 þ bx þ c is sometimes called a quadratic function.

12.17 Obtain the graph of y ¼ x3 þ 2x2  7x  3. SOLUTION x

24

23

22

21

0

1

2

3

y

27

9

11

5

23

27

21

21

The graph is shown in Fig. 12-18. Points marked  are not listed in the table; they were added in order to improve the accuracy of the graph. Point A is called a relative maximum point; it is not the highest point on the entire curve, but points on either side are lower. Point B is called a relative minimum point. The calculus enables us to determine such relative maximum and minimum points.

Fig. 12-17

Fig. 12-18

12.18 Obtain the graph of x2 þ y2 ¼ 36.

SOLUTION

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi We may write y2 ¼ 36  x2 or y ¼ + 36  x2 . Note that x must have a value between 26 and þ6 if y is to be a real number. x

26

25

24

23

22

21

0

1

2

3

4

5

6

y

0

pffiffiffiffiffi + 11

pffiffiffiffiffi + 20

pffiffiffiffiffi + 27

pffiffiffiffiffi + 32

pffiffiffiffiffi + 35

+6

pffiffiffiffiffi + 35

pffiffiffiffiffi + 32

pffiffiffiffiffi + 27

pffiffiffiffiffi + 20

pffiffiffiffiffi + 11

0

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi The points to be plotted are (6, 0), (5, 11), (5,  11), (4, 20), (4; 20), etc. Figure 12-19 shows the graph, a circle of radius 6. In general, the graph of x2 þ y2 ¼ a2 is a circle with center at the origin and radius a. It should be noted that if the units had not been taken as the same on the x and y axes, the graph would not have looked like a circle.

102

FUNCTIONS AND GRAPHS

[CHAP. 12

Fig. 12-19

12.19 Determine whether the graph is symmetric with respect to the y axis, x axis, or the origin. (a) y ¼ 4x (b) x2 þ y2 ¼ 8

(c) xy2 ¼ 1 (d) x ¼ y2 þ 1

(e) y ¼ xp3 ffiffiffi (f) y ¼ x

SOLUTION (a) Origin (b) y axis x axis Origin (c) x axis (d ) x axis (e) Origin ( f ) None

Since y ¼ 4(x) is equivalent to y ¼ 4x. Since (x)2 þ y2 ¼ 8 is equivalent to x2 þ y2 ¼ 8. Since x2 þ (y)2 ¼ 8 is equivalent to x2 þ y2 ¼ 8. Since (x)2 þ (y)2 ¼ 8 is equivalent to x2 þ y2 ¼ 8. Since x(y)2 ¼ 1 is equivalent to xy2 ¼ 1. Since x ¼ (y)2 þ 1 is equivalent to x ¼ y2 þ 1. Since (y) ¼ (x)3 is equivalent to y ¼ x3 .

12.20 Use the graph of y ¼ x3 to graph y ¼ x3 þ 1. SOLUTION The graph of y ¼ x3 is shown in Fig. 12-20. The graph of y ¼ x3 þ 1 is the graph of y ¼ x3 1 unit up and is shown in Fig. 12-21.

Fig. 12-20

Fig. 12-21

CHAP. 12]

103

FUNCTIONS AND GRAPHS

Fig. 12-22

Fig. 12-23

12.21 Use the graph of y ¼ j xj to graph y ¼ j x þ 2j. SOLUTION The graph of y ¼ jxj is shown in Fig. 12-22. The graph of y ¼ j x þ 2j is the graph of y ¼ jxj shifted 2 units to the left, since j x þ 2j ¼ jx  (2)j and is shown in Fig. 12-23.

12.22 Use the graph of y ¼ x2 to graph y ¼ x2 . SOLUTION The graph of y ¼ x2 is shown in Fig. 12-24. The graph of y ¼ x2 is the graph of y ¼ x2 reflected across the x axis and is shown in Fig. 12-25.

Fig. 12-24

Fig. 12-25

12.23 A man has 40 ft of wire fencing with which to form a rectangular garden. The fencing is to be used only on three sides of the garden, his house providing the fourth side. Determine the maximum area which can be enclosed. SOLUTION Let x ¼ length of each of two of the fenced sides of the rectangle; then 40 2 2x ¼ length of the third fenced side.

104

FUNCTIONS AND GRAPHS

[CHAP. 12

The area A of the garden is A ¼ x(40  2x) ¼ 40x  2x2 . We wish to find the maximum value of A. A table of values and the graph of A plotted against x are shown. It is clear that x must have a value between 0 and 20 ft if A is to be positive. x

0

5

8

10

12

15

20

A

0

150

192

200

192

150

0

From the graph in Fig. 12-26 the maximum point P has coordinates (10, 200) so that the dimensions of the garden are 10 ft by 20 ft and the area is 200 ft2 .

Fig. 12-26

12.24 A rectangular piece of tin has dimensions 12 in. by 18 in. It is desired to make an open box from this material by cutting out equal squares from the corners and then bending up the sides. What are the dimensions of the squares cut out if the volume of the box is to be as large as possible?

SOLUTION Let x be the length of the side of the square cut out from each corner. The volume V of the box thus obtained is V ¼ x(12  2x)(18  2x). It is clear that x must be between 0 and 6 in. if there is to be a box (see Fig. 12-27). x V

0 0

1 160

2

2 12

224

227 12

3

3 12

4

5

6

216

192 12

160

80

0

From the graph, the value of x corresponding to the maximum value of V lies between 2 and 2.5 in. By plotting more points it is seen that x ¼ 2.4 in. approximately. Problems such as this and Problem 12.23 may often be solved easily and exactly by methods of the calculus.

Fig. 12-27

CHAP. 12]

105

FUNCTIONS AND GRAPHS

12.25 A cylindrical can is to have a volume of 200 cubic inches. Find the dimensions of the can which is made of the least amount of material. SOLUTION Let x be the radius and y the height of the cylinder. The area of the top or bottom of the can is px2 and the lateral area is 2pxy; then the total area S ¼ 2px2 þ 2pxy. The volume of the cylinder is px2 y, so that px2 y ¼ 200 and y ¼ 200=px2 . Then  S ¼ 2px2 þ 2px

200 px2

 or

S ¼ 2px2 þ

400 : x

A table of values and the graph of S plotted against x (Fig. 12-28) are shown. We take p ¼ 3:14 approximately. x

1

2

3

3.2

3.5

4

4.5

5

6

7

8

S

406

225

190

189

191

200

216

237

293

365

452

From the graph in Fig. 12-28, minimum S ¼ 189 in2 occurs when x ¼ 3.2 in. approximately; and from y ¼ 200=px2 we have y ¼ 6.2 in. approximately.

Fig. 12-28

12.26 Find approximately the values of x for which x3 þ 2x2  7x  3 ¼ 0. SOLUTION Consider y ¼ x3 þ 2x2  7x  3: We must find values of x for which y ¼ 0. From the graph of y ¼ x3 þ 2x2  7x  3, which is shown in Fig. 12-18, it is clear that there are three real values of x for which y ¼ 0 (the values of x where the curve intersects the x axis). These values are x ¼ 3:7, x ¼ 0:4, and x ¼ 2:1 approximately. More exact values may be obtained by advanced techniques.

12.27 The following table shows the population of the United States (in millions) for the years 1840, 1850, . . . , 1950. Graph these data.

Year

1840

1850

1860

1870

1880

1890

1900

1910

1920

1930

1940

1950

Population (millions)

17.1

23.2

31.4

39.8

50.2

62.9

76.0

92.0

105.7

122.8

131.7

150.7

106

FUNCTIONS AND GRAPHS

[CHAP. 12

SOLUTION See Fig. 12-29.

Fig. 12-29

Fig. 12-30

12.28 The time T (in seconds) required for one complete vibration of a simple pendulum of length l (in centimeters) is given by the following observations obtained in a physics laboratory. Exhibit graphically T as a function of l.

l

16.2

22.2

33.8

42.0

53.4

66.7

74.5

86.6 100.0

T

0.81

0.95

1.17

1.30

1.47

1.65

1.74

1.87

2.01

SOLUTION The observation points are connected by a smooth curve (Fig. 12-30) as is usually done in science and engineering.

Supplementary Problems 12.29

A rectangle has sides of lengths x and 2x. Express the area A of the rectangle as a function of its (a) side x, (b) perimeter P, and (c) diagonal D.

12.30

Express the area A of a circle in terms of its (a) radius r, (b) diameter d, and (c) circumference, C.

CHAP. 12]

107

FUNCTIONS AND GRAPHS

12.31

Express the area A of an isosceles triangle as a function of x and y, where x is the length of the two equal sides and y is the length of the third side.

12.32

The side of a cube has length x. Express (a) x as a function of the volume V of the cube, (b) the surface area S of the cube as a function of x, and (c) the volume V as a function of the surface S.

12.33

Given y ¼ 5 þ 3x  2x2 , find the values of y corresponding to x ¼ 3, 2, 1, 0, 1, 2, 3:

12.34

Extend the table of values in Problem 12.33 by finding the values of y which correspond to x ¼ 5=2, 3=2, 1=2, 1=2, 3=2, 5=2:

12.35

State the domain and range for each equation. (a) y ¼ 2x þ 3 (b) y ¼ x2  5 (c) y ¼ x3  4

12.36

For which of these relations is y a function of x? (a) y ¼ x3 þ 2

(d ) y ¼ x2  5

(b) x ¼ y þ 2

(e) x2 þ y2 ¼ 5 pffiffiffiffiffiffiffiffiffiffiffi (f ) y ¼ + x  7

3

(c) x ¼ y2 þ 4 12.37

p ffiffiffiffiffiffiffiffiffiffiffiffiffi 3 1  2x x (h) y ¼ xþ1 4 (i) y ¼ x (g) y ¼

(d) y ¼ 5  2x2 2 (e) y ¼ xþ6 pffiffiffiffiffiffiffiffiffiffiffi (f ) y ¼ x  5

(g) x ¼ jyj ffiffiffiffiffiffiffiffiffiffiffi p 3 xþ1 pffiffiffiffiffiffiffiffiffiffiffi (i) y ¼ 5 þ x (h) y ¼

If f (x) ¼ 2x2 þ 6x  1, find f (3), f (2), f (0), f (1=2), f (3).

12.39

u2  2u , find (a) F(1), (b) F(2), (c) F(x), 1þu x1 , find If G(x) ¼ xþ1   x G(x þ h)  G(x) , (c) G(x2 þ 1). (a) G , (b) xþ1 h

12.40

If F(x, y) ¼ 2x2 þ 4xy  y2 , find (a) F(1, 2),

12.41

Plot the following points on a rectangular coordinate system:

12.38

If F(u) ¼

(a) (1, 3),

(b) (22, 1),

(d ) F(x).

(b) F(2, 3),

(c) (1=2, 2),

(c) F(x þ 1, y 1):

(d ) (3, 2=3),

pffiffiffi (e) (  3, 3).

12.42

If y ¼ 3x þ 2, (a) obtain the values of y corresponding to x ¼ 22, 21, 0, 1, 2 and (b) plot the points (x, y) thus obtained.

12.43

Determine whether the graph of each relation is symmetric with respect to the y axis, x axis, or origin. (a) y ¼ 2x4 þ 3

(b) y ¼ (xp ffiffiffiffiffiffiffiffiffiffi 3) ffi (c) y ¼  9  x 3

12.44

(g) y2 ¼ x þ 2

(e) y ¼ 5x ( f ) y ¼ 7x2 þ 4 3

(h) y ¼ 3x  1 (i) y ¼ 5x

State how the graph of the first equation relates to the graph of the second equation. (a) (b) (c) (d ) (e)

12.45

(d ) y ¼ 3

y ¼ x4 and y ¼ x4 y ¼ 3x and y ¼ x y ¼ x2 þ 10 and y ¼ x2 y ¼ (x  1)3 and y ¼ x3 y ¼ x2  7 and y ¼ x2

(f) (g) (h) (i) ( j)

Graph the functions (a) f (x) ¼ 1 2 2x,

y ¼ jxj þ 1 and y ¼ jxj y ¼ jx þ 5j and y ¼ jxj y ¼ x3 and y ¼ x3 y ¼ x2 =6 and y ¼ x2 y ¼ (x þ 8)2 and y ¼ x2 (b) f(x) ¼ x 2 2 4x þ 3, (c) f(x) ¼ 4 2 3x 2 x 2.

108

FUNCTIONS AND GRAPHS

[CHAP. 12

12.46

Graph y ¼ x 3 2 6x 2 þ 11x 2 6.

12.47

Graph (a) x 2 þ y 2 ¼ 16,

12.48

There is available 120 ft of wire fencing with which to enclose two equal rectangular gardens A and B, as shown in Fig. 12-31. If no wire fencing is used along the sides formed by the house, determine the maximum combined area of the gardens.

12.49

Find the area of the largest rectangle which can be inscribed in a right triangle whose legs are 6 and 8 in. respectively (see Fig. 12-32).

(b) x 2 þ 4y 2 ¼ 16.

Fig. 12-31

Fig. 12-32

12.50

Obtain the relative maximum and minimum values of the function f (x) ¼ 2x 3 2 15x 2 þ 36x 2 23.

12.51

From the graph of y ¼ x 3 2 7x þ 6 obtain the roots of the equation x 3 2 7x þ 6 ¼ 0.

12.52

Show that the equation x 3 2 x 2 þ 2x 2 3 ¼ 0 has only one real root.

12.53

Show that x 4 2 x 2 þ 1 ¼ 0 cannot have any real roots.

12.54

The percentage of workers in the USA employed in agriculture during the years 1860, 1870, . . . , 1950 is given in the following table. Graph the data.

12.55

Year

1860

1870

1880

1890

1900

1910

1920

1930

1940

1950

% of all workers in agriculture

58.9

53.0

49.4

42.6

37.5

31.0

27.0

21.4

18.0

12.8

The total time required to bring an automobile to a stop after perceiving danger is composed of the reaction time (time between recognition of danger and application of brakes) plus braking time (time for stopping after application of brakes). The following table gives the stopping distances d (feet) of an automobile traveling at speeds v (miles per hour) at the instant danger is sighted. Graph d against v.

Speed v (mi/hr)

20

30

40

50

60

70

Stopping distance d (ft)

54

90

138

206

292

396

CHAP. 12]

12.56

109

FUNCTIONS AND GRAPHS

The time t taken for an object to fall freely from rest through various heights h is given in the following table.

Time t (sec)

1

2

3

4

5

6

Height h (ft)

16

64

144

256

400

576

(a) Graph h against t. (b) How long would it take an object to fall freely from rest through 48 ft? 300 ft? (c) Through what distance can an object fall freely from rest in 3.6 seconds? ANSWERS TO SUPPLEMENTARY PROBLEMS 12.29 12.30 12.31 12.32

P2 2D2 , A¼ 18 5 2 2 p d C , A¼ A¼ A ¼ pr 2 , 4 4p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y A¼ x2  y2 =4 ¼ 4x2  y2 2 4 rffiffiffiffiffiffiffiffi p ffiffiffiffi S3 S pffiffiffiffiffi 3 2 x ¼ V, 6S S ¼ 6x , ¼ V¼ 216 36 A¼

A ¼ 2x2 ,

12.33

x

23

22

21

0

1

2

3

y

222

29

0

5

6

3

24

12.34

x

5=2

3=2

1=2

1=2

3=2

5=2

y

215

24

3

6

5

0

Domain ¼ fall real numbersg; Domain ¼ fall real numbersg; Domain ¼ fall real numbersg; Domain ¼ fall real numbersg; Domain ¼ fall real numbers=26g; Domain ¼ fall real numbers 5g; Domain ¼ fall real numbersg; Domain ¼ fall real numbers=21g; Domain ¼ fall real numbers=0g;

12.35

(a) (b) (c) (d ) (e) (f ) (g) (h) (i)

12.36

(a) Function (b) Function (c) Not a function

12.37

f (3) ¼ 1,

12.38

(a) 1=2,

(d ) Function (e) Not a function ( f ) Not a function

f (2) ¼ 5,

(b) 0,

(c)

f (0) ¼ 1,

x2  2x ; 1þx

(d )

range ¼ fall real numbersg range ¼ fall real numbers 25g range ¼ fall real numbersg range ¼ fall real numbers 5g range ¼ fall real numbers=0g range ¼ fall real numbers 0g range ¼ fall real numbersg range ¼ fall real numbers=1g range ¼ fall real numbers=0g (g) Not a function (h) Function (i) Function f (1=2) ¼ 5=2, x2 þ 2x 1x

f (3) ¼ 35

110

FUNCTIONS AND GRAPHS

1 , 2x þ 1

(b)

2 , (x þ 1)(x þ h þ 1)

12.39

(a)

12.40

(a) 6,

12.41

See Fig. 12-33.

12.42

(a) 24, 21, 2, 5, 8

(c)

x2 x2 þ 2

(c) 2x 2 þ 4xy 2 y 2 þ 6y 2 3

(b) 23,

(b) See Fig. 12-34.

Fig. 12-34

Fig. 12-33

12.43

(a) y axis (b) Origin (c) None

12.44

(a) (b) (c) (d ) (e)

[CHAP. 12

(d ) y axis (e) Origin ( f ) y axis

(g) x axis (h) None (i) Origin

Reflected across x axis y increases 3 times as fast Shifted 10 units up Shifted 1 unit right Shifted 7 units down

Fig. 12-35

(f ) (g) (h) (i) ( j)

Shifted 1 unit up Shifted 5 units left Reflected across the x axis y increases 1/6 as fast Shifted 8 units left

Fig. 12-36

CHAP. 12]

111

FUNCTIONS AND GRAPHS

12.45

(a) See Fig. 12-35.

(b) See Fig. 12-36.

(c) See Fig. 12-37.

12.46

See Fig. 12-38.

12.47

(a) See Fig. 12-39.

12.48

1200 ft2

12.49

12 in.2

12.50

Maximum value of f(x) is 5 (at x ¼ 2); minimum value of f(x) is 4 (at x ¼ 3).

(b) See Fig. 12-40.

Fig. 12-37

Fig. 12-39

Fig. 12-38

Fig. 12-40

112

FUNCTIONS AND GRAPHS

Fig. 12-41

[CHAP. 12

Fig. 12-42

Fig. 12-43

Fig. 12-44

Fig. 12-45

CHAP. 12]

FUNCTIONS AND GRAPHS

12.51

Roots are x ¼ 23, x ¼ 1, x ¼ 2.

12.52

See Fig. 12-41.

12.53

See Fig. 12-42.

12.54

See Fig. 12-43.

12.55

See Fig. 12-44.

12.56

(a) See Fig. 12-45;

(b) 1.7 sec, 4.3 sec;

(c) 207 ft

113

CHAPTER 13

Linear Equations in One Variable 13.1

LINEAR EQUATIONS

A linear equation in one variable has the form ax þ b ¼ 0, where a = 0 and b are constants. The solution of this equation is given by x ¼ b=a. When a linear equation is not in the form ax þ b ¼ 0, we simplify the equation by multiplying each term by the LCD for all fractions in the equation, removing any parentheses, or combining like terms. In some equations we do more than one of the procedures. EXAMPLE 13.1.

Solve the equation x þ 8  2(x þ 1) ¼ 3x  6 for x. x þ 8  2(x þ 1) ¼ 3x  6 First we remove the parentheses. x þ 8  2x  2 ¼ 3x  6 We now combine like terms. x þ 6 ¼ 3x  6 Now get the variable terms on one side of the equation by x þ 6  3x ¼ 3x  6  3x subtracting 3x from each side of the equation. 4x þ 6 ¼ 6 Now we subtract 6 from each side of the equation to get the 4x þ 6  6 ¼ 6  6 variable term on one side of the equation by itself. 4x ¼ 12 Finally we divide each side of the equation by the coefficient of 4x 12 the variable, which is 24. ¼ 4 4 x¼3 Now we check this solution in the original equation.

Check: 3 þ 8  2(3 þ 1) ? 3(3)  6 11  2(4) ? 9  6 11  8 ? 3 3¼3

13.2

The question mark indicates that we don’t know for sure that the two quantities are equal. The solution checks.

LITERAL EQUATIONS

Most literal equations we encounter are formulas. Frequently, we want to use the formula to determine a value other than the standard one. To do this, we consider all variables except the one we are interested in as constants and solve the equation for the desired variable. 114

CHAP. 13]

EXAMPLE 13.2.

13.3

LINEAR EQUATIONS IN ONE VARIABLE

Solve p ¼ 2(lþ w) for l. p ¼ 2(l þ w) p ¼ 2l þ 2w p  2w ¼ 2l ( p  2w)=2 ¼ l l ¼ ( p  2w)=2

115

Remove the parentheses first. Subtract 2w from each side of the equation. Divide by 2, the coefficient of l. Rewrite the equation. Now we have a formula that can be used to determine l.

WORD PROBLEMS

In solving a word problem, the first step is to decide what is to be found. The next step is to translate the conditions stated in the problem into an equation or to state a formula that expresses the conditions of the problem. The solution of the equation is the next step. EXAMPLE 13.3. If the perimeter of a rectangle is 68 meters and the length is 14 meters more than the width, what are the dimensions of the rectangle? Let w ¼ the number of meters in the width and w þ 14 ¼ the number of meters in the length. 2½(w þ 14) þ w ¼ 68 2w þ 28 þ 2w ¼ 68 4w þ 28 ¼ 68 4w ¼ 40 w ¼ 10 w þ 14 ¼ 24 The rectangle is 24 meters long by 10 meters wide. EXAMPLE 13.4. The sum of two numbers is 24 and their difference is 6. What are the numbers? Let n ¼ the smaller number and n þ 6 ¼ the larger number. n þ (n þ 6) ¼ 4 n þ n þ 6 ¼ 4 2n þ 6 ¼ 4 2n ¼ 10 n ¼ 5 nþ6¼1 The two numbers are 25 and 1. EXAMPLE 13.5. If one pump can fill a pool in 16 hours and if two pumps can fill the pool in 6 hours, how fast can the second pump fill the pool? Let h ¼ the numbers of hours for the second pump to fill the pool. 1 1 1 þ ¼ h 16 6     1 1 1 ¼ 48h 48h þ h 16 6 48 þ 3h ¼ 8h 48 ¼ 5h 9:6 ¼ h The second pump takes 9.6 hours (or 9 hours and 36 minutes) to fill the pool.

116

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

EXAMPLE 13.6. How many liters of pure alcohol must be added to 15 liters of a 60% alcohol solution to obtain an 80% alcohol solution? Let n ¼ the number of liters of pure alcohol to be added. n þ 0:60(15) ¼ 0:80(n þ 15)

(The sum of the amount of alcohol in each quantity is equal to the amount of alcohol in the mixture:)

n þ 9 ¼ 0:8n þ 12 0:2n ¼ 3 n ¼ 15 Fifteen liters of pure alcohol must be added.

Solved Problems 13.1

Solve each of the following equations. (a) x þ 1 ¼ 5, x ¼ 5  1, x ¼ 4. Check: Put x ¼ 4 in the original equation and obtain 4 þ l ? 5, 5 ¼ 5. (b) 3x  7 ¼ 14, 3x ¼ 14 þ 7, 3x ¼ 21, x ¼ 7. Check: 3(7) 27 ? 14, 14 ¼ 14. (c) 3x þ 2 ¼ 6x  4, 3x  6x ¼ 4  2, 3x ¼ 6, x ¼ 2. (d ) x þ 3(x  2) ¼ 2x  4, x þ 3x  6 ¼ 2x  4, 4x  2x ¼ 6  4, 2x ¼ 2, x ¼ l. (e) 3x  2 ¼ 7  2x, 3x þ 2x ¼ 7 þ 2, 5x ¼ 9, x ¼ 9/5. ( f ) 2(t þ 3) ¼ 5(t  1)  7(t  3), 2t þ 6 ¼ 5t  5  7t þ 21, 4t ¼ 10, t ¼ 10/4 ¼ 5/2. (g) 3x þ 4(x  2) ¼ x  5 þ 3(2x  1), 3x þ 4x  8 ¼ x  5 þ 6x  3, 7x  8 ¼ 7x  8. This is an identity and is true for all values of x. (h)

x  3 2x þ 4 ¼ , 5(x  3) ¼ 2(2x þ 4), 5x  15 ¼ 4x þ 8, x ¼ 23. 2 5

(i) 3 þ 2½ y  (2y þ 2) ¼ 2½ y þ (3y  1), 3 þ 2½ y  2y  2 ¼ 2½ y þ 3y  1, 3 þ 2y  4y  4 ¼ 2y þ 6y  2, 2y  1 ¼ 8y  2, 10y ¼ 1, y ¼ 1=10: ( j ) (s þ 3)2 ¼ (s  2)2  5, s2 þ 6s þ 9 ¼ s2  4s þ 4  5, 6s þ 4s ¼ 9  1, s ¼ 1: x2 x4 (k) ¼ , (x  2)(x þ 4) ¼ (x  4)(x þ 2), x2 þ 2x  8 ¼ x2  2x  8, 4x ¼ 0, x ¼ 0. xþ2 xþ4 Check:

02 04 ? , 0þ2 0þ4

1 ¼ 1:

3x þ 1 3x  2 ¼ , (x þ 1)(3x þ 1) ¼ (x þ 2)(3x  2), 3x2 þ 4x þ 1 ¼ 3x2 þ 4x  4 or 1 ¼ 24. xþ2 xþ1 There is no value of x which satisfies this equation. 5 5 (m) þ ¼ 6. Multiplying by 2x, 5(2) þ 5 ¼ 12x, 12x ¼ 15, x ¼ 5/4. x 2x xþ3 5 1 þ ¼ . Multiplying by 2x(x  1), the LCD of the fractions, (n) 2x x1 2 (l )

(x þ 3)(x  1) þ 5(2x) ¼ x(x  1),

x2 þ 2x  3 þ 10x ¼ x2  x,

13x ¼ 3,

x ¼ 3=13:

CHAP. 13]

(o)

2 4 16  ¼ 2 . Multiplying by (x  3)(x þ 3) or x2  9, x3 xþ3 x 9 2(x þ 3)  4(x  3) ¼ 16,

( p)

1 1 1 1  ¼  , y yþ3 yþ2 yþ5 ( y þ 2)( y þ 5) ¼ y( y þ 3),

(q)

2x þ 6  4x þ 12 ¼ 16,

2x ¼ 2,

( y þ 3)  y ( y þ 5)  ( y þ 2) ¼ , y( y þ 3) ( y þ 2)( y þ 5) y2 þ 7y þ 10 ¼ y2 þ 3y,

3 2 9  ¼ x2  4x 2x2  5x  12 2x2 þ 3x

or

x ¼ 1: 3 3 ¼ , y( y þ 3) ( y þ 2)( y þ 5)

4y ¼ 10, y ¼ 5=2:

3 2 9  ¼ : x(x  4) (2x þ 3)(x  4) x(2x þ 3)

Multiplying by x(x  4)(2x þ 3), the LCD of the fractions, 3(2x þ 3)  2x ¼ 9(x  4), 6x þ 9  2x ¼ 9x  36, 13.2

117

LINEAR EQUATIONS IN ONE VARIABLE

45 ¼ 5x,

x ¼ 9:

Solve for x. (a) 2x  4p ¼ 3x þ 2p, 2x  3x ¼ 2p þ 4p, x ¼ 6p, x ¼ 6p: ba (b) ax þ a ¼ bx þ b, ax  bx ¼ b  a, x(a  b) ¼ b  a, x ¼ ¼ 1 provided a = b. ab If a ¼ b the equation is an identity and is true for all values of x. 4b  4d 4b þ 4d ¼ provided 3a = 2c. 2c  3a 3a  2c If 3a ¼ 2c there is no solution unless d ¼ 2b, in which case the original equation is an identity. 3x þ a 4x þ b b2  a2 ¼ , 3ax þ a2 ¼ 4bx þ b2 , 3ax  4bx ¼ b2  a2 , x ¼ (provided 3a = 4b). (d ) b a 3a  4b

(c) 2cx þ 4d ¼ 3ax  4b, 2cx  3ax ¼ 4b  4d, x ¼

13.3

Express each statement in terms of algebraic symbols. (a) One more than twice a certain number. Let x ¼ the number. Then 2x ¼ twice the number, and one more than twice the number ¼ 2x þ 1. (b) Three less than five times a certain number. Let x ¼ the number. Then three less than five times the number ¼ 5x 2 3. (c) Each of two numbers whose sum is 100. If x ¼ one of the numbers, then 100 2 x ¼ the other number. (d ) Three consecutive integers (for example, 5, 6, 7). If x is the smallest integer, then (x þ 1) and (x þ 2) are the other two integers. (e) Each of two numbers whose difference is 10. Let x ¼ the smaller number; then (x þ 10) ¼ the larger number. ( f ) The amount by which 100 exceeds three times a given number. Let x ¼ given number. Then the excess of 100 over 3x is (100 2 3x). (g) Any odd integer. Let x ¼ any integer. Then 2x is always an even integer, and (2x þ 1) is an odd integer. (h) Four consecutive odd integers (for example, 1, 3, 5, 7; 17, 19, 21, 23). The difference between two consecutive odd integers is 2. Let 2x þ 1 ¼ smallest odd integer. Then the required numbers are 2x þ 1, 2x þ 3, 2x þ 5, 2x þ 7. (i) The number of cents in x dollars. Since 1 dollar ¼ 100 cents, x dollars ¼ 100x cents. ( j ) John is twice as old as Mary, and Mary is three times as old as Bill. Express each of their ages in terms of a single unknown.

118

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

Let x ¼ Bill’s age. Then Mary’s age is 3x and John’s age is 2(3x) ¼ 6x.   Another method. Let y ¼ John’s age. Then Mary’s age ¼ 12 y and Bill’s age ¼ 13 12 y ¼ 16 y: (k) The three angles A, B, C of a triangle if angle A has 108 more than twice the number of degrees in angle C. Let C ¼ x8; then A ¼ (2x þ 10)8. Since A þ B þ C ¼ 1808, B ¼ 1808 2 (A þ C) ¼ (170 – 3x)8. (l ) The time it takes a boat traveling at a speed of 20 mi/hr to cover a distance of x miles. Distance ¼ speed  time. Then time ¼

distance x mi x ¼ ¼ hr: speed 20 mi/ hr 20

(m) The perimeter and area of a rectangle if one side is 4 ft longer than twice the other side. Let x ft ¼ length of shorter side; then (2x þ 4)ft ¼ length of longer side. The perimeter ¼ 2(x) þ 2(2x þ 4) ¼ (6x þ 8)ft, and the area ¼ x(2x þ 4) ft2 : (n) The fraction whose numerator is 3 less than 4 times its denominator. Let x ¼ denominator; then numerator ¼ 4x 2 3. The fraction is (4x  3)=x. (o) The number of quarts of alcohol contained in a tank holding x gallons of a mixture which is 40% alcohol by volume. In x gallons of mixture are 0.40x gallons of alcohol or 4(0.40x) ¼ 1.6x quarts of alcohol. 13.4

The sum of two numbers is 21, and one number is twice the other. Find the numbers. SOLUTION Let x and 2x be the two numbers. Then x þ 2x ¼ 21 or x ¼ 7, and the required numbers are x ¼ 7 and 2x ¼ 14. Check. 7 þ 14 ¼ 21 and 14 ¼ 2(7), as required.

13.5

Ten less than four times a certain number is 14. Determine the number. SOLUTION Let x ¼ required number. Then 4x  10 ¼ 14, 4x ¼ 24, and x ¼ 6. Check. Ten less than four times 6 is 4(6) 210 ¼ 14, as required.

13.6

The sum of three consecutive integers is 24. Find the integers. SOLUTION Let the three consecutive integers be x, x þ 1, x þ 2. Then x þ (x þ 1) þ (x þ 2) ¼ 24 or x ¼ 7, and the required integers are 7, 8, 9.

13.7

The sum of two numbers is 37. If the larger is divided by the smaller, the quotient is 3 and the remainder is 5. Find the numbers. SOLUTION Let x ¼ smaller number, 37 2 x ¼ larger number. Then

larger number 5 ¼3þ smaller number smaller number

or

37  x 5 ¼3þ : x x

Solving, 37  x ¼ 3x þ 5, 4x ¼ 32, x ¼ 8. The required numbers are 8, 29.

13.8

A man is 41 years old and his son is 9. In how many years will the father be three times as old as the son?

CHAP. 13]

119

LINEAR EQUATIONS IN ONE VARIABLE

SOLUTION Let x ¼ required number of years. Father’s age in x years ¼ 3(son’s age in x years) 41 þ x ¼ 3(9 þ x)

13.9

and

x ¼ 7 years:

Ten years ago Jane was four times as old as Bianca. Now she is only twice as old as Bianca. Find their present ages. SOLUTION Let x ¼ Bianca’s present age; then 2x ¼ Jane’s present age. Jane’s age ten years ago ¼ 4(Bianca’s age ten years ago) 2x  10 ¼ 4(x  10) and x ¼ 15 years: Hence Bianca’s present age is x ¼ 15 years and Jane’s present age is 2x ¼ 30 years. Check. Ten years ago Bianca was 5 and Jane 20, i.e., Jane was four times as old as Bianca.

13.10 Robert has 50 coins, all in nickels and dimes, amounting to $3.50. How many nickels does he have? SOLUTION Let x = number of nickels; then 50 2 x = number of dimes. Amount in nickels þ amount in dimes ¼ 350¢ 5x¢ þ 10(50  x)¢ ¼ 350¢

from which x ¼ 30 nickels:

13.11 In a purse are nickels, dimes, and quarters amounting to $1.85. There are twice as many dimes as quarters, and the number of nickels is two less than twice the number of dimes. Determine the number of coins of each kind. SOLUTION Let x ¼ number of quarters; then 2x ¼ no. of dimes, and 2(2x)  2 ¼ 4x  2 ¼ no. of nickels. Amount in quarters þ amount in dimes þ amount in nickels ¼ 185¢ 25(x)¢

þ

10(2x)¢

þ

5(4x  2)¢

¼ 185¢

from which x ¼ 3:

Hence there are x ¼ 3 quarters, 2x ¼ 6 dimes, and 4x 2 2 ¼ 10 nickels. Check. 3 quarters ¼ 75¢, 6 dimes ¼ 60¢. 10 nickels ¼ 50¢, and their sum ¼ $1.85.

13.12 The tens digit of a certain two-digit number exceeds the units digit by 4 and is 1 less than twice the units digit. Find the two-digit number. SOLUTION Let x ¼ units digit; then x þ 4 ¼ tens digit. Since the tens digit ¼ 2(units digit) 2 1, we have x þ 4 ¼ 2(x) 2 1 or x ¼ 5. Thus x ¼ 5, x þ 4 ¼ 9, and the required number is 95.

120

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

13.13 The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 4/7 times the original number. Determine the original number. SOLUTION Let x ¼ units digit; 12 2 x ¼ tens digit. Original number ¼ 10(12 2 x) þ x; reversing digits, the new number ¼ 10x þ (12 2 x). Then 4 4 new number ¼ (original number) or 10x þ (12  x) ¼ ½10(12  x) þ x. 7 7 Solving, x ¼ 4, 12 2 x ¼ 8, and the original number is 84.

13.14 A man has $4000 invested, part at 5% and the remainder at 3% simple interest. The total income per year from these investments is $168. How much does he have invested at each rate? SOLUTION Let x ¼ amount invested at 5%; $4000 2 x ¼ amount at 3%. Interest from 5% investment þ interest from 3% investment ¼ $168 0:05x þ 0:03(4000  x) ¼ 168: Solving, x ¼ $2400 at 5%, $4000 2 x ¼ $1600 at 3%.

13.15 What amount should an employee receive as bonus so that she would net $500 after deducting 30% for taxes? SOLUTION Let x = required amount. Then or

required amount  taxes ¼ $500 x  0:30x ¼ $500

and x ¼ $714:29:

13.16 At what price should a merchant mark a sofa that costs $120 in order that it may be offered at a discount of 20% on the marked price and still make a profit of 25% on the selling price? SOLUTION Let x ¼ marked price; then sale price ¼ x 2 0.20x ¼ 0.80x. Since profit ¼ 25% of sale price, cost ¼ 75% of sale price. Then cost ¼ 0:75(sale price) $120 ¼ 0:75(0:8x), $120 ¼ 0:6x

and

x ¼ $200:

13.17 When each side of a given square is increased by 4 feet the area is increased by 64 square feet. Determine the dimensions of the original square. SOLUTION Let x ¼ side of given square; x þ 4 ¼ side of new square. New area ¼ old area þ 64 (x þ 4)2 ¼ x2 þ 64

from which x ¼ 6 ft:

CHAP. 13]

LINEAR EQUATIONS IN ONE VARIABLE

121

13.18 One leg of a right triangle is 20 inches and the hypotenuse is 10 inches longer than the other leg. Find the lengths of the unknown sides. SOLUTION Let x ¼ length of unknown leg; x þ 10 ¼ length of hypotenuse. Square of hypotenuse ¼ sum of squares of legs (x þ 10)2 ¼ x2 þ (20)2

from which x ¼ 15 in:

The required sides are x ¼ 15 in. and x þ 10 ¼ 25 in.

13.19 Temperature Fahrenheit ¼ 95(temperature Celsius) þ 32. At what temperature have the Fahrenheit and Celsius readings the same value? SOLUTION Let x ¼ required temperature ¼ temperature Fahrenheit ¼ temperature Celsius. Then x ¼ 95 x þ 32 or x ¼ 408. Thus 408F ¼ 408C.

13.20 A mixture of 40 lb of candy worth 60¢ a pound is to be made up by taking some worth 45¢/lb and some worth 85¢/lb. How many pounds of each should be taken? SOLUTION Let x ¼ weight of 45¢ candy; 40 2 x ¼ weight of 85¢ candy.

or

Value of 45¢=lb candy þ value of 85¢=lb candy ¼ value of mixture x(45¢) þ (40  x)(85¢) ¼ 40(60¢):

Solving, x ¼ 25 lb of 45¢/lb candy; 40 2 x ¼ 15 lb of 85¢/lb candy.

13.21 A tank contains 20 gallons of a mixture of alcohol and water which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will be 25% alcohol by volume? SOLUTION Let x ¼ volume of 40% solution to be removed. or

Volume of alcohol in final solution ¼ volume of alcohol in 20 gal of 25% solution 0:40(20  x) ¼ 0:25(20) from which x ¼ 7:5 gallons:

13.22 What weight of water must be evaporated from 40 lb of a 20% salt solution to produce a 50% solution? All percentages are by weight. SOLUTION Let x ¼ weight of water to be evaporated. or

Weight of salt in 20% solution ¼ weight of salt in 50% solution 0:20(40 lb) ¼ 0:50(40 lb  x) from which x ¼ 24 lb:

13.23 How many quarts of a 60% alcohol solution must be added to 40 quarts of a 20% alcohol solution to obtain a mixture which is 30% alcohol? All percentages are by volume.

122

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

SOLUTION Let x ¼ number of quarts of 60% alcohol to be added. Alcohol in 60% solution þ alcohol in 20% solution ¼ alcohol in 30% solution or

þ

0:60x

0:20(40)

¼ 0:30(x þ 40)

and

x ¼ 13 13 qt:

13.24 Two unblended manganese (Mn) ores contain 40% and 25% of manganese respectively. How many tons of each must be mixed to give 100 tons of blended ore containing 35% of manganese? All percentages are by weight. SOLUTION Let x ¼ weight of 40% ore required; 100 2 x ¼ weight of 25% ore required. Mn from 40% ore þ Mn from 25% ore ¼ total Mn in 100 tons of mixture 0:40x

þ

0:25(100  x) ¼ 0:35(100)

from which x ¼ 66 23 tons of 40% ore and 100  x ¼ 33 13 tons of 25% ore.

13.25 Two cars A and B having average speeds of 30 and 40 mi/hr respectively are 280 miles apart. They start moving toward each other at 3:00 P.M. At what time and where will they meet? SOLUTION Let t ¼ time in hours each car travels before they meet. Distance ¼ speed  time. Distance traveled by A þ distance traveled by B ¼ 280 miles 30t

þ

40t

¼ 280

from which t ¼ 4 hr.

They meet at 7:00 P.M. at a distance 30t ¼ 120 mi from initial position of A or at a distance 40t ¼ 160 mi from initial position of B.

13.26 A and B start from a given point and travel on a straight road at average speeds of 30 and 50 mi/hr respectively. If B starts 3 hr after A, find (a) the time and (b) the distance they travel before meeting. SOLUTION Let t and (t 2 3) be the number of hours A and B respectively travel before meeting. (a) Distance in miles ¼ average speed in mi/hr  time in hours. When they meet, distance covered by A ¼ distance covered by B 30t ¼ 50(t  3) from which t ¼ 7 12 hr. Hence A travels t ¼ 7 12 hr and B travels (t  3) ¼ 4 12 hr. (b) Distance ¼ 30t ¼ 30(7 12 ) ¼ 225 mi, or distance ¼ 50(t  3) ¼ 50(4 12 ) ¼ 225 mi.

13.27 A and B can run around a circular mile track in 6 and 10 minutes respectively. If they start at the same instant from the same place, in how many minutes will they pass each other if they run around the track (a) in the same direction, (b) in opposite directions? SOLUTION Let t ¼ required time in minutes. (a) They will pass each other when A covers 1 mile more than B. The speeds A and B are 1/6 and 1/10 mi/min respectively. Then, since distance ¼ speed  time: Distance by A  distance by B ¼ 1 mile

CHAP. 13]

LINEAR EQUATIONS IN ONE VARIABLE

123

1 1 t  t ¼ 1 and t ¼ 15 minutes. 6 10 (b) Distance by A þ distance by B ¼ 1 mile 1 1 t þ t ¼ 1 and t ¼ 15=4 minutes. 6 10

13.28 A boat, propelled to move at 25 mi/hr in still water, travels 4.2 mi against the river current in the same time that it can travel 5.8 mi with the current. Find the speed of the current. SOLUTION Let v ¼ speed of current. Then, since time ¼ distance/speed, time against the current ¼ time in direction of the current 4:2 mi 5:8 mi ¼ and v ¼ 4 mi/hr. (25  v) mi/hr (25 þ v) mi/hr

or

13.29 A can do a job in 3 days, and B can do the same job in 6 days. How long will it take them if they work together? SOLUTION Let n ¼ number of days it will take them working together. In 1 day A does 1/3 of the job and B does 1/6 of the job, thus together completing 1/n of the job (in 1 day). Then 1 1 1 þ ¼ 3 6 n

from which n ¼ 2 days:

Another method. In n days A and B together complete   1 1 þ ¼ 1 complete job. n 3 6

Solving, n ¼ 2 days:

13.30 A tank can be filled by three pipes separately in 20, 30, and 60 minutes respectively. In how many minutes can it be filled by the three pipes acting together? SOLUTION Let t ¼ time required, in minutes. 1 1 1 In 1 minute three pipes together fill ( 20 þ 30 þ 60 ) of the tank. Then in t minutes they together fill 

 1 1 1 t þ þ ¼ 1 complete tank: 20 30 60

Solving, t ¼ 10 minutes:

13.31 A and B working together can complete a job in 6 days. A works twice as fast as B. How many days would it take each of them, working alone, to complete the job? SOLUTION Let n, 2n ¼ number of days required by A and B respectively, working alone, to do the job. In 1 day A can do l/n of the job and B can do l/2n of the job. Then in 6 days they can do   1 1 þ ¼ 1 complete job: 6 n 2n

Solving, n ¼ 9 days, 2n ¼ 18 days:

124

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

13.32 A’s rate of doing work is three times that of B. On a given day A and B work together for 4 hours; then B is called away and A finishes the rest of the job in 2 hours. How long would it take B to do the complete job alone? SOLUTION Let t, 3t ¼ time in hours required by A and B respectively, working alone, to do the job. In 1 hour A does 1/t of job and B does l/3t of job. Then     1 1 1 þ2 ¼ 1 complete job: 4 þ t 3t t

Solving, 3t ¼ 22 hours:

13.33 A man is paid $18 for each day he works and forfeits $3 for each day he is idle. If at the end of 40 days he nets $531, how many days was he idle? SOLUTION Let x ¼ number of days idle; 40 2 x ¼ number of days worked. Amount earned  amount forfeited ¼ $531 or $18(40  x)  3x ¼ $531 and x ¼ 9 days idle.

Supplementary Problems 13.34

Solve each of the following equations. (a) 3x  2 ¼ 7

(h) (2x þ 1)2 ¼ (x  1)2 þ 3x(x þ 2)

(b) y þ 3(y  4) ¼ 4

(i)

(c) 4x  3 ¼ 5  2x

3 4 1  ¼ z 5z 10

2x þ 1 x  4 þ ¼3 x xþ1 5 5 2 2 (k)  ¼  y1 yþ1 y2 yþ3 7 2 4 þ ¼ (l) 2 x  4 x2  3x þ 2 x2 þ x  2

(d) x  3  2(6  2x) ¼ 2(2x  5)

( j)

2t  9 3t þ 4 ¼ 3 2 2x þ 3 x  1 (f ) ¼ 2x  4 x þ 1

(e)

(g) (x  3)2 þ (x þ 1)2 ¼ (x  2)2 þ (x þ 3)2 13.35

Solve for the indicated letter. (a) 2(x  p) ¼ 3(6p  x) : x (b) 2by  2a ¼ ay  4b : y 2x  a 2x  b (c) ¼ :x b a

13.36

(d)

xa xc ¼ :x xb xd

(e)

1 1 1 þ ¼ :y ay by c

Express each of the following statements in terms of algebraic symbols. (a) Two more than five times a certain number. (b) Six less than twice a certain number. (c) Each of two numbers whose difference is 25. (d ) The squares of three consecutive integers. (e) The amount by which five times a certain number exceeds 40. ( f ) The square of any odd integer. (g) The excess of the square of a number over twice the number.

CHAP. 13]

LINEAR EQUATIONS IN ONE VARIABLE

125

(h) The number of pints in x gallons. (i)

The difference between the squares of two consecutive even integers.

( j) Bob is six years older than Jane who is half as old as Jack. Express each of their ages in terms of a single unknown. (k) The three angles A, B, C of a triangle ABC if angle A exceeds twice angle B by 208. (l)

The perimeter and area of a rectangle if one side is 3 ft shorter than three times the other side.

(m) The fraction whose denominator is 4 more than twice the square of the numerator. (n) The amount of salt in a tank holding x quarts of water if the concentration is 2 lb of salt per gallon.

13.37

(a) One half of a certain number is 10 more than one sixth of the number. Find the number. (b) The difference between two numbers is 20 and their sum is 48. Find the numbers. (c) Find two consecutive even integers such that twice the smaller exceeds the larger by 18. (d ) The sum of two numbers is 36. If the larger is divided by the smaller, the quotient is 2 and the remainder is 3. Find the numbers. (e) Find two consecutive positive odd integers such that the difference of their squares is 64. ( f ) The first of three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.

13.38

(a) A father is 24 years older than his son. In 8 years he will be twice as old as his son. Determine their present ages. (b) Mary is fifteen years older than her sister Jane. Six years ago Mary was six times as old as Jane. Find their present ages. (c) Larry is now twice as old as Bill. Five years ago Larry was three times as old as Bill. Find their present ages.

13.39

(a) In a purse is $3.05 in nickels and dimes, 19 more nickels than dimes. How many coins are there of each kind? (b) Richard has twice as many dimes as quarters, amounting to $6.75 in all. How many coins does he have? (c) Admission tickets to a theater were 60¢ for adults and 25¢ for children. Receipts for the day showed that 280 persons attended and $140 was collected. How many children attended that day?

13.40

(a) The tens digit of a certain two-digit number exceeds the units digit by 3. The sum of the digits is 1/7 of the number. Find the number. (b) The sum of the digits of a certain two-digit number is 10. If the digits are reversed, a new number is formed which is one less than twice the original number. Find the original number. (c) The tens digit of a certain two-digit number is 1/3 of the units digit. When the digits are reversed, the new number exceeds twice the original number by 2 more than the sum of the digits. Find the original number.

13.41

(a) Goods cost a merchant $72. At what price should he mark them so that he may sell them at a discount of 10% from his marked price and still make a profit of 20% on the selling price? (b) A woman is paid $20 for each day she works and forfeits $5 for each day she is idle. At the end of 25 days she nets $450. How many days did she work? (c) A labor report states that in a certain factory a total of 400 men and women are employed. The average daily wage is $16 for a man and $12 for a woman. If the labor cost is $5720 per day, how many women are employed? (d ) A woman has $450 invested, part at 2% and the remainder at 3% simple interest. How much is invested at each rate if the total annual income from these investments is $11? (e) A man has $2000 invested at 7% and $5000 at 4% simple interest. What additional sum must he invest at 6% to give him an overall return of 5%?

126

13.42

LINEAR EQUATIONS IN ONE VARIABLE

[CHAP. 13

(a) The perimeter of a rectangle is 110 ft. Find the dimensions if the length is 5 ft less than twice the width. (b) The length of a rectangular floor is 8 ft greater than its width. If each dimension is increased by 2 ft, the area is increased by 60 ft2. Find the dimensions of the floor. (c) The area of a square exceeds the area of a rectangle by 3 in2. The width of the rectangle is 3 in. shorter and the length 4 in. longer than the side of the square. Find the side of the square. (d ) A piece of wire 40 in. long is bent into the form of a right triangle, one of whose legs is 15 in. long. Determine the lengths of the other two sides. (e) The length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4 ft wide. If the area of the walk is 784 ft2, determine the dimensions of the pool.

13.43

(a) Lubricating oil worth 28 cents/quart is to be mixed with oil worth 33 cents/quart to make up 45 quarts of a mixture to sell at 30¢/qt. What volume of each grade should be taken? (b) What weight of water must be added to 50 lb of a 36% sulfuric acid solution to yield a 20% solution? All percentages are by weight. (c) How many quarts of pure alcohol must be added to 10 quarts of a 15% alcohol solution to obtain a mixture which is 25% alcohol? All percentages are by volume. (d ) There is available 60 gallons of a 50% solution of glycerin and water. What volume of water must be added to the solution to reduce the glycerin concentration to 12%? All percentages are by volume. (e) The radiator of a jeep has a capacity of 4 gallons. It is filled with an anti-freeze solution of water and glycol which analyzes 10% glycol. What volume of the mixture must be drawn off and replaced with glycol to obtain a 25% glycol solution? All percentages are by volume. ( f ) One thousand quarts of milk testing 4% butterfat are to be reduced to 3%. How many quarts of cream testing 23% butterfat must be separated from the milk to produce the required result? All percentages are by volume. (g) There are available 10 tons of coal containing 2.5% sulfur, and also supplies of coal containing 0.80% and 1.10% sulfur respectively. How many tons of each of the latter should be mixed with the original 10 tons to give 20 tons containing 1.7% sulfur?

13.44

(a) Two motorists start toward each other at 4:30 P.M. from towns 255 miles apart. If their respective average speeds are 40 and 45 mi/hr, at what time will they meet? (b) Two planes start from Chicago at the same time and fly in opposite directions, one averaging a speed of 40 mi/hr greater than the other. If they are 2000 miles apart after 5 hours, find their average speeds. (c) At what rate must motorist A travel to overtake motorist B who is traveling at a rate 20 mi/hr slower if A starts two hours after B and wishes to overtake B in 4 hours? (d ) A motorist starts from city A at 2:00 P.M. and travels to city B at an average speed of 30 mi/hr. After resting at B for one hour, she returns over the same route at an average speed of 40 mi/hr and arrives at A that evening at 6:30 P.M. Determine the distance between A and B. (e) Tom traveled a distance of 265 miles. He drove at 40 mi/hr during the first part of the trip and at 35 mi/hr during the remaining part. If he made the trip in 7 hours, how long did he travel at 40 mi/hr? ( f ) A boat can move at 8 mi/hr in still water. If it can travel 20 miles downstream in the same time it can travel 12 miles upstream, determine the rate of the stream. (g) The speed of a plane is 120 mi/hr in a calm. With the wind it can cover a certain distance in 4 hours, but against the wind it can cover only 3/5 of that distance in the same time. Find the velocity of the wind.

13.45

(a) A farmer can plow a certain field three times as fast as his son. Working together, it would take them 6 hours to plow the field. How long would it take each to do it alone? (b) A painter can do a given job in 6 hours. Her helper can do the same job in 10 hours. The painter begins the work and after 2 hours is joined by the helper. In how many hours will they complete the job? (c) One group of workers can do a job in 8 days. After this group has worked 3 days, another group joins it and together they complete the job in 3 more days. In what time could the second group have done the job alone?

CHAP. 13]

127

LINEAR EQUATIONS IN ONE VARIABLE

(d ) A tank can be filled by two pipes separately in 10 and 15 minutes respectively. When a third pipe is used simultaneously with the first two pipes, the tank can be filled in 4 minutes. How long would it take the third pipe alone to fill the tank? ANSWERS TO SUPPLEMENTARY PROBLEMS 13.34

13.35

(a) x ¼ 3

(d ) x ¼ 5

(g) x ¼ 1=2

( j) x ¼ 1=4

(b) y ¼ 4

(e) t ¼ 6

(h) all values of x (identity)

(k) y ¼ 5

(c) x ¼ 4=3

( f ) x ¼ 1=11

(i) z ¼ 22

(l) x ¼ 1

(a) x ¼ 4p

(c) x ¼

(b) y ¼ 2 if a = 2b 13.36

aþb 2

if a = b

(d ) x ¼

bc  ad bþcad

(e) y ¼

(a) 5x þ 2

( j) Jane’s age x, Bob’s age x þ 6, Jack’s age 2x

(b) 2x  6

(k) B ¼ x8, A ¼ (2x þ 20)8, C ¼ (160  3x)8

(c) x þ 25, x

(l ) One side is x, adjacent side is 3x  3.

(d ) x2 , (x þ 1)2 , (x þ 2)2

Perimeter ¼ 8x  6, area ¼ 3x2  3x x (m) 2 2x þ 4

(e) 5x  40 ( f ) (2x þ 1)2 where x ¼ integer (g) x2  2x

(n)

(h) 8x

x lb salt 2

(i) (2x þ 2)2  (2x)2 , x ¼ integer 13.37

(a) 30

(b) 34, 14

(c) 20, 22

13.38

(a) Father 40, son 16

(b) Mary 24, Jane 9

13.39

(a) 14 dimes, 33 nickels

(b) 15 quarters, 30 dimes

(c) 200 adults, 80 children

13.40

(a) 63

(b) 37

(c) 26

13.41

(a) $100

13.42

(a) width 20 ft, length 35 ft

(d ) other leg 8 ft, hypotenuse 17 ft

(b) width 10 ft, length 18 ft

(e) 30 ft by 60 ft

(b) 23 days

(d) 25, 11

(e) 15, 17

( f ) 16, 6, 32

(c) Larry 20, Bill 10

(c) 170 women (d) $200 at 3%, $250 at 2%

(e) $1000

(c) 9 in. 13.43

(a) 18 qt of 33¢, 27 qt of 28¢

(e) 2/3 gal

(b) 40 lb

( f ) 50 qt

(c) 4/3 qt

(g) 6.7 tons of 0.80%, 3.3 tons of 1.10%

(d ) 190 gal 13.44

13.45

(a) 7:30 P.M.

(c) 60 mi/hr

(e) 4 hr

(b) 180, 220 mi/hr

(d ) 60 mi

( f ) 2 mi/hr

(a) Father 8 hr, son 24 hr

(b) 2 12 hr

(c) 12 days

(g) 30 mi/hr

(d ) 12 minutes

ac þ bc ab

CHAPTER 14

Equations of Lines 14.1

SLOPE OF A LINE

The equation ax þ by ¼ c, where not both a and b are 0 and a, b, and c are real numbers is the standard (or general) form of the equation of a line. The slope of a line is the ratio of the change in y compared to the change in x. slope ¼

change in y change in x

If (x1 , y1 ) and (x2 , y2 ) are two points on a line and m is the slope of the line, then m¼ EXAMPLE 14.1.

y2  y1 x2  x1

when

x2 = x1 :

What is the slope of the line through the points (5, 28) and (6, 2)? m¼

y2  y1 2  (8) 10 ¼ ¼ 10 ¼ x2  x1 56 1

The slope of the line through the two points (5, 28) and (6, 2) is 210. EXAMPLE 14.2. What is the slope of the line 3x  4y ¼ 12? First we need to find two points that satisfy the equation of the line 3x  4y ¼ 12: If x ¼ 0, then 3(0) 2 4y ¼ 12 and y ¼ 23. Thus, one point is (0, 3). If x ¼ 24, then 3(24) 2 4y ¼ 12 and y ¼ 26. So, (24, 26) is another point on the line. m¼

y2  y1 3  (6) 3 ¼ ¼ x2  x1 0  (4) 4

The slope of the line 3x  4y ¼ 12 is 3=4:

In Example 14.1, the slope of the line is negative. This means that, as we view the graph of the line from left to right, as x increases y decreases (see Fig. 14-1). In Example 14.2, the slope is positive, which means that as x increases so does y (see Fig. 14-2). A horizontal line y ¼ k, where k is a constant, has zero slope. Since all of the y values are the same, y2  y1 ¼ 0. 128

CHAP. 14]

129

EQUATIONS OF LINES

Fig. 14-1

Fig. 14-2

A vertical line x ¼ k, where k is a constant, does not have a slope, that is, the slope is not defined. Since all of the x values are the same, x2  x1 ¼ 0 and division by zero is not defined.

14.2

PARALLEL AND PERPENDICULAR LINES

Two non-vertical lines are parallel if and only if the slopes of the lines are equal. EXAMPLE 14.3. Show that the figure PQRS with vertices P(0, 22), Q(22, 3), R(3, 5), and S(5, 0) is a parallelogram. Quadrilateral PQRS is a parallelogram if PQ and RS are parallel and PS and QR are parallel. 3  (2) 5 5 ¼ ¼ 2  0 2 2 0  (2) 2 slope (PS) ¼ ¼ and 50 5

05 5 ¼ 53 2 53 2 slope (QR) ¼ ¼ 3  (2) 5

slope (PQ) ¼

and

slope (RS) ¼

Since PQ and RS have the same slope, they are parallel, and since PS and QR have the same slope, they are parallel. Thus, the opposite sides of PQRS are parallel, so PQRS is a parallelogram.

EXAMPLE 14.4. Show that the points A(0, 4), B(2, 3), and C(4, 2) are collinear, that is, lie on the same line. The points A, B, and C are collinear if the slopes of the lines through any two pairings of the points are the same. slope (AB) ¼

34 1 ¼ 20 2

and

slope (BC) ¼

23 1 ¼ 42 2

The lines AB and BC have the same slope and share a common point, B, so the lines are the same line. Thus, the points A, B, and C are collinear.

Two non-vertical lines are perpendicular if and only if the product of their slopes is 21. The slope of each line is said to be the negative reciprocal of the slope of the other line.

130

EQUATIONS OF LINES

[CHAP. 14

EXAMPLE 14.5. Show that the line through the points A(3, 3) and B(6, 23) is perpendicular to the line through the points C(4, 2) and D(8, 4). slope (AB) ¼

3  3 6 ¼ ¼ 2 63 3

and

slope (CD) ¼

42 2 1 ¼ ¼ 84 4 2

Since (22)(1/2) ¼ 21, the lines AB and CD are perpendicular.

14.3

SLOPE –INTERCEPT FORM OF EQUATION OF A LINE

If a line has slope m and y intercept (0, b), then for any point (x, y), where x = 0, on the line we have m¼

yb x0

and

y ¼ mx þ b:

The slope– intercept form of the equation of a line with slope m and y intercept b is y ¼ mx þ b. EXAMPLE 14.6. Find the slope and y intercept of the line 3x þ 2y ¼ 12. We solve the equation 3x þ 2y ¼ 12 for y to get y ¼  32 x þ 6: The slope of the line is  32 and the y intercept is 6. EXAMPLE 14.7. Find the equation of the line with slope 24 and y intercept 6. The slope of the line is 24, so m ¼ 24 and the y intercept is 6, so b ¼ 6. Substituting into y ¼ mx þ b, we get y ¼ 4x þ 6 for the equation of the line.

14.4

SLOPE –POINT FORM OF EQUATION OF A LINE

If a line has slope m and goes through a point (x1 , y1 ), then for any other point (x, y) on the line, we have m ¼ (y  y1 )=(x  x1 ) and y  y1 ¼ m(x  x1 ): The slope– point form of the equation of a line is y  y1 ¼ m(x  x1 ): EXAMPLE 14.8. Write the equation of the line passing through the point (1, 22) and having slope 22/3. Since (x1 , y1 ) ¼ (1, 2) and m ¼ 2=3, we substitute into y  y1 ¼ m(x  x1 ) to get y þ 2 ¼ 2=3(x  1). Simplifying we get 3(y þ 2) ¼ 2(x  1), and finally 2x þ 3y ¼ 4: The equation of the line through (1, 2) with slope 2=3 is 2x þ 3y ¼ 4:

14.5

TWO-POINT FORM OF EQUATION OF A LINE

If a line goes through the points (x1 , y1 ) and (x2 , y2 ), it has slope m ¼ (y2  y1 )=(x2  x1 ) if x2 = x1 : Substituting into the equation y  y1 ¼ m(x  x1 ), we get y  y1 ¼

y2  y1 (x  x1 ): x2  x1

The two-point form of the equation of a line is y  y1 ¼

y2  y1 (x  x1 ) if x2 =x1 : x2  x1

If x2 ¼ x1 , we get the vertical line x ¼ x1 . If y2 ¼ y1 , we get the horizontal line y ¼ y1 :

CHAP. 14]

131

EQUATIONS OF LINES

EXAMPLE 14.9. Write the equation of the line passing through (3, 6) and (24, 4). Let (x1 , y1 ) ¼ (3, 6) and (x2 , y2 ) ¼ (4, 4) and substitute into y  y1 ¼

y2  y1 (x  x1 ): x2  x1

y6¼

46 (x  3) 4  3

7(y  6) ¼ 2(x  3) 7y þ 42 ¼ 2x þ 6 2x  7y ¼ 36 The equation of the line through the points (3, 6) and (24, 4) is 2x  7y ¼ 36:

14.6

INTERCEPT FORM OF EQUATION OF A LINE

If a line has x intercept a and y intercept b, it goes through the points (a, 0) and (0, b). The equation of the line is yb¼

0b (x  0) a0

if a = 0,

which simplifies to bx þ ay ¼ ab: If both a and b are non-zero we get x=a þ y=b ¼ 1: If a line has x intercept a and y intercept b and both a and b are non-zero, the equation of the line is x y þ ¼ 1: a b EXAMPLE 14.10. Find the intercepts of the line 4x  3y ¼ 12: We divide the equation 4x  3y ¼ 12 by 12 to get x y þ ¼ 1: 3 4 The x intercept is 3 and the y intercept is 24 for the line 4x  3y ¼ 12: EXAMPLE 14.11. Write the equation of the line that has an x intercept of 2 and a y intercept of 5. We have a ¼ 2 and b ¼ 5 for the equation x y þ ¼ 1: a b Substituting we get

x y þ ¼ 1: 2 5

Simplifying we get

5x þ 2y ¼ 10:

The line with x intercept 2 and y intercept 5 is 5x þ 2y ¼ 10:

Solved Problems 14.1

What is the slope of the line through each pair of points? (a) (4, 1) and (7, 6)

(b) (3, 9) and (7, 4)

(c) (4, 1) and (4, 3)

(d ) (23, 2) and (2, 2)

132

EQUATIONS OF LINES

[CHAP. 14

SOLUTION 61 5 ¼ 74 3 49 5 (b) m ¼ ¼ 73 4 31 2 (c) m ¼ ¼ 4  (4) 0 22 0 ¼ ¼0 (d ) m ¼ 2  (3) 5 (a) m ¼

14.2

The slope of the line is 5/3. The slope of the line is 25/4. Slope is not defined for this line. The slope of the line is 0.

Determine whether the line containing the points A and B is parallel, perpendicular, or neither to the line containing the points C and D. (a) A(2, 4), B(3, 8), C(5, 1), and D(4, 23) (b) A(2, 23), B(24, 5), C(0, 21), and D(24, 24) (c) A(1, 9), B(4, 0), C(0, 6), and D(5, 3) (d ) A(8, 21), B(2, 3), C(5, 1), and D(2, 21) SOLUTION 84 4 3  1 4 ¼ ¼ 4; slope (CD) ¼ ¼ ¼4 32 1 45 1 Since the slopes are equal, the lines AB and CD are parallel. 5  (3) 8 4 4  (1) 3 3 ¼ ¼ ; slope (CD) ¼ ¼ ¼ (b) slope (AB) ¼ 4  2 6 3 4  0 4 4 Since (24/3)(3/4) ¼ 21, the lines AB and CD are perpendicular. 0  9 9 36 3 (c) slope (AB) ¼ ¼ ¼ 3; slope (CD) ¼ ¼ 41 3 50 5 Since the slopes are not equal and do not have a product of 21, the lines AB and CD are neither parallel nor perpendicular. 3  (1) 4 2 1  1 2 2 (d ) slope (AB) ¼ ¼ ¼ ; slope (CD) ¼ ¼ ¼ 28 6 3 25 3 3 Since the slopes are not equal and do not have a product of 21, the lines AB and CD are neither parallel nor perpendicular. (a) slope (AB) ¼

14.3

Determine whether the given three points are collinear or not. (a) (0, 3), (1, 1), and (2, 21) SOLUTION 1  3 2 (a) m1 ¼ ¼ ¼ 2 10 1

(b) (1, 5), (22, 21), and (23, 24)

and

m2 ¼

1  1 2 ¼ ¼ 2 21 1

Since the line between (0, 3) and (1, 1) and the line between (1, 1) and (2, 21) have the same slope, the points (0, 3), (1, 1), and (2, 21) are collinear. (b) m1 ¼

1  5 6 ¼ ¼ 2 2  1 3

and

m2 ¼

4  (1) 3 ¼ ¼3 3  (2) 1

Since the slope of the line between (1, 5) and (22, 21) and the slope of the line between (22, 21) and (23, 24) are different, the points (1, 5), (22, 21), and (23, 24) are not collinear.

14.4

Write the equation of the line with slope m and y intercept b. (a) m ¼ 2=3, b ¼ 6

(b) m ¼ 3, b ¼  4

(c) m ¼ 0, b ¼ 8

(d ) m ¼ 3, b ¼ 0

CHAP. 14]

133

EQUATIONS OF LINES

SOLUTION (a) (b) (c) (d )

14.5

y ¼ mx þ b ¼ 22/3x þ 6 y ¼ mx þ b ¼ 23x 2 4 y ¼ mx þ b ¼ 0x þ 8 y ¼ mx þ b ¼ 3x þ 0

2x þ 3y ¼ 18 3x þ y ¼ 24 y¼8 3x 2 y ¼ 0

Write the equation of the line that contains point P and has slope m. (a) P(2, 5), m ¼ 4

(b) P(1, 4), m ¼ 0

(c) P(21, 26), m ¼ 1/4

(d ) P(2, 23), m ¼ 23/7

SOLUTION

We use the formula y  y1 ¼ m(x  x1 ). (a) (b) (c) (d )

14.6

y 2 5 ¼ 4(x 2 2) y ¼ 4 ¼ 0(x 2 1) y 2 (26) ¼ l/4(x 2 (21)) y 2 (23) ¼ 23/7(x 22)

4x 2 y ¼ 3 y¼4 x 2 4y ¼ 23 3x þ 7y ¼ 215

Write the equation of the line passing through points P and Q. (a) P(1, 24), Q(2, 3) (b) P(6, 21), Q(0, 2)

(c) P(2l, 4), Q(3, 4) (d) P(1, 5), Q(22, 3)

(e) P(7, 1), Q(8, 3) ( f ) P(4, 21), Q(4, 3)

SOLUTION 3  (4) (x  2) 21 2  (1) (x  0) y2¼ 06 44 y4¼ (x  3) 3  (1) 35 y3¼ (x  (2)) 2  1 31 (x  8) y3¼ 87

(a) y  3 ¼

y  3 ¼ 7(x  2)

7x  y ¼ 11

(b)

y  2 ¼ 1=2x

x þ 2y ¼ 4

y  4 ¼ 0(x  3)

y¼4

y  3 ¼ 2=3(x þ 2)

2x  3y ¼ 13

y  3 ¼ 2(x  8)

2x  y ¼ 13

(c) (d ) (e)

( f ) Since P and Q have the same x value, slope is not defined. However, the line through P and Q has to have 4 for its x coordinate in all points. Thus, the line is x ¼ 4.

14.7

Write the equation of the line that has x intercept 23 and y intercept 4. SOLUTION x y þ ¼1 a b

14.8

so

x y þ ¼1 3 4

4x  3y ¼ 12

Write the equation of the line through (25, 6) that is parallel to the line 3x 2 4y ¼ 5. SOLUTION We write the equation 3x 2 4y ¼ 5 in slope– intercept form to identify its slope, y ¼ 3/4x 2 5/4. Since the form is y ¼ mx þ b, m ¼ 3/4. Parallel lines have the same slope, so the line we want has a slope of 3/4. Now that we have the slope and a point the line goes through, so we can write the equation using the point – slope form: y  y1 ¼ m(x  x1 ). Substituting we get y 2 6 ¼ 3/4(x þ 5). Simplifying we get 4y 2 24 ¼ 3x þ 15 and finally 3x 2 4y ¼ 239. The equation we want is 3x 2 4y ¼ 239.

134

14.9

EQUATIONS OF LINES

[CHAP. 14

Write the equation of the line through (4, 6) that is perpendicular to the line 2x 2 y ¼ 8. SOLUTION In the slope – intercept form the given line is y ¼ 2x 2 8. The slope of the line is 2, so the slope of a perpendicular line is the negative reciprocal of 2, which is 21/2. We want to write the equation of the line with slope of 21/2 and going through (4, 6). Thus, y 2 6 ¼ 2l/2(x 2 4), so 2y 2 12 ¼ 2x þ 4, and finally x þ 2y ¼ 16. The line we want is x þ 2y ¼ 16.

Supplementary Problems 14.10

What is the slope of the line through each pair of points? (a) (21, 2), (4, 23) (b) (3, 4), (24, 23)

14.11

P(4, 2), Q(8, 3), R(22, 8), and S(1, 24) P(0, 25), Q(15, 0), R(1, 2), and S(0, 5) P(27, 8), Q(8, 27), R(28, 10), and S(6, 24) P(8, 22), Q(2, 8), R(22, 28), and S(28, 22)

Determine a constant real number k such that the lines AB and CD are (1) parallel and (2) perpendicular. (a) (b) (c) (d )

14.13

A(2, 1), B(6, 3), C(4, k), and D(3, 1) A(1, k), B(2, 3), C(1, 7), and D(3, 6) A(9, 4), B(k, 10), C(11, 22), and D(22, 4) A(1, 2), B(4, 0), C(k, 2), and D(1, 23)

Determine whether the given three points are collinear or not. (a) (23, 1), (211, 21), and (215, 22)

14.14

(c) m ¼ 2/3, b ¼ 22 (d ) m ¼ 4, b ¼ 0

(e) P(2, 6), m ¼ 25 ( f ) P(21, 6), m ¼ 0

(d ) P(10, 2), Q(5, 2) (e) P(3, 6), Q(23, 8) ( f ) P(24, 2), Q(2, 4)

(g) P(21, 3), Q(0, 6) (h) P(0, 0), Q(23, 6)

Write the equation of the line that has x intercept a and y intercept b. (a) a ¼ 22, b ¼ 22

14.18

(c) P(4, 21), m ¼ 2/3 (d ) P(0, 4), m ¼ 24/3

Write the equation of the line through the points P and Q. (a) P(1, 2), Q(2, 4) (b) P(1.6, 3), Q(0.3, 1.4) (c) P(0.7, 3), Q(0.7, 23)

14.17

(e) m ¼ 21/2, b ¼ 3 ( f ) m ¼ 25/6, b ¼ 1/6

Write the equation of the line that goes through point P and has slope m. (a) P(25, 2), m ¼ 21 (b) P(24, 23), m ¼ 4

14.16

(b) (1, 1), (4, 2), and (2, 3)

Write the equation of the line with slope m and y intercept b. (a) m ¼ 23, b ¼ 4 (b) m ¼ 0, b ¼ 23

14.15

(e) (21, 5), (22, 3) ( f ) (7, 3), (8, 23)

Determine whether the line containing the points P and Q is parallel, perpendicular, or neither to the line containing the points R and S. (a) (b) (c) (d )

14.12

(c) (5, 4), (5, 22) (d ) (25, 3), (2, 3)

(b) a ¼ 6, b ¼ 23

(c) a ¼ 21/2, b ¼ 4

Write the equation of the line through point P and parallel to the line l. (a) P(2, 24), line l: y ¼ 4x 2 6 (b) P(1, 0), line l: y ¼ 3x þ 1

(c) P(21, 21), line l: 4x þ 5y ¼ 5 (d ) P(3, 5), line l: 3x 2 2y ¼ 18

(d ) a ¼ 6, b ¼ 1/3

CHAP. 14]

14.19

Write the equation of the line through the point P and perpendicular to the line l. (a) P(2, 21), line l: x ¼ 4y (b) P(0, 6), line l: 2x þ 3y ¼ 5

14.20

(c) P(1, 1), line l: 3x 2 2y ¼ 4 (d ) P(1, 22), line l: 4x þ y ¼ 7

Determine if the triangle with vertices A, B, and C is a right triangle. (a) A(4, 0), B(7, 27), and C(2, 25) (b) A(5, 8), B(22, 1), and C(2, 23)

14.21

(c) A(2, 1), B(3, 21), and C(1, 22) (d ) A(26, 3), B(3, 25), and C(21, 5)

Show by using slopes that the diagonals PR and QS of quadrilateral PQRS are perpendicular. (a) P(0, 0), Q(5, 0), R(8, 4), and S(3, 4)

14.22

135

EQUATIONS OF LINES

(b) P(23, 0), Q(6, 23), R(7, 5), and S(3, 3)

Show that the points P, Q, R, and S are the vertices of a parallelogram PQRS. (a) P(5, 0), Q(8, 2), R(6, 5), and S(3, 3)

(b) P(29, 0), Q(210, 26), R(4, 8), and S(5, 14)

14.23

Write the equation of the line through (7, 3) that is parallel to the x axis.

14.24

Write the equation of the horizontal line through the point (22, 23).

14.25

Write the equation of the vertical line through the point (2, 4).

14.26

Write the equation of the line through (5, 8) that is perpendicular to the x axis.

ANSWERS TO SUPPLEMENTARY PROBLEMS 14.10

(a) 21

14.11

(a) perpendicular, slopes 1/4 and 24 (b) perpendicular, slopes 1/3 and 23

14.12

(a) (1) 3/2 (2) 21 (b) (1) 7/2 (2) 1

14.13

(a) yes: m ¼ 1/4

14.14

(a) y ¼ 23x þ 4 (b) y ¼ 23

14.15

(a) x þ y ¼ 23 (b) 4x 2 y ¼ 213

14.16

(a) y ¼ 2x (b) 80x 2 65y ¼ 267

14.17

(a) x þ y ¼ 22

(b) x 2 2y ¼ 6

(c) 8x 2 y ¼ 24

14.18

(a) y ¼ 4x 2 12

(b) y ¼ 3x 2 3

(c) 4x þ 5y ¼ 29

14.19

(a) 4x þ y ¼ 7

14.20

(a) yes AC ? BC

14.21

(a) yes: slopes are 1/2 and 22

(b) 1

(c) not defined

(d) 0

(e) 2

( f ) 26

(c) parallel, slopes 21 and 21 (d ) neither, slopes 25/3 and 21

(c) (1) 24 (2) 153/13 (d ) (1) 213/2 (2) 13/3 (b) no: slopes are not equal (c) 2x 2 3y ¼ 6 (d) y ¼ 4x (c) 2x 2 3y ¼ 11 (d ) 4x þ 3y ¼ 12 (c) 10x ¼ 7 (d ) y ¼ 2

(b) 3x 2 2y ¼ 212 (b) yes AB ? BC

(e) x þ 2y ¼ 6 ( f ) 5x þ 6y ¼ 1 (e) 5x þ y ¼ 16 (f) y ¼ 6 (e) x þ 3y ¼ 21 ( f ) x 2 3y ¼ 210

(c) 2x þ 3y ¼ 5 (c) yes AB ? BC

(g) 3x 2 y ¼ 26 (h) y ¼ 22x

(d ) x þ 18y ¼ 6 (d ) 3x 2 2y ¼ 21 (d ) x 2 4y ¼ 9 (d ) yes AC ? BC

(b) yes: slopes are 1/2 and 22

136

EQUATIONS OF LINES

14.22

(a) yes: PQ k RS and QR k SP

14.23

y¼3

14.24

y ¼ 23

14.25

x¼2

14.26

x¼5

(b) yes: PQ k RS and QR k SP

[CHAP. 14

CHAPTER 15

Simultaneous Linear Equations 15.1

SYSTEMS OF TWO LINEAR EQUATIONS

A linear equation in two variables x and y is of the form ax þ by ¼ c where a, b, c are constants and a, b are not both zero. If we consider two such equations a1 x þ b 1 y ¼ c 1 a2 x þ b2 y ¼ c 2 we say that we have two simultaneous linear equations in two unknowns or a system of two linear equations in two unknowns. A pair of values for x and y, (x, y), which satisfies both equations is called a simultaneous solution of the given equations. Thus the simultaneous solution of x þ y ¼ 7 and x 2 y ¼ 3 is (5, 2). Three methods of solving such systems of linear equations are illustrated here. A.

Solution by addition or subtraction. If necessary, multiply the given equations by such numbers as will make the coefficients of one unknown in the resulting equations numerically equal. If the signs of the equal coefficients are unlike, add the resulting equations; if like, subtract them. Consider (1) 2x  y ¼ 4 (2) x þ 2y ¼ 3: To eliminate y, multiply (1) by 2 and add to (2) to obtain 2  (1): 4x  2y ¼ 8 (2): x þ 2y ¼ 3 Addition: 5x

¼5

or

x ¼ 1:

Substitute x ¼ 1 in (1) and obtain 2 2 y ¼ 4 or y ¼ 22. Thus the simultaneous solution of (1) and (2) is (1, 2). Check: Put x ¼ 1, y ¼ 22 in (2) and obtain 1 þ 2(22) ? 23, 23, ¼ 23. 137

138

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

B.

Solution by substitution. Find the value of one unknown in either of the given equations and substitute this value in the other equation. For example, consider the system (1), (2) above. From (1) obtain y ¼ 2x24 and substitute this value into (2) to get x þ 2(2x 2 4) ¼ 23 which reduces to x ¼ 1. Then put x ¼ 1 into either (1) or (2) and obtain y ¼ 22. The solution is (1, 22). C. Graphical solution. Graph both equations, obtaining two straight lines. The simultaneous solution is given by the coordinates (x, y) of the point of intersection of these lines. Fig. 15-1 shows that the simultaneous solution of (1) 2x 2 y ¼ 4 and (2) x þ 2y ¼ 23 is x ¼ 1, y ¼ 22, also written (1, 22). If the lines are parallel, the equations are inconsistent and have no simultaneous solution. For example, (3) x þ y ¼ 2 and (4) 2x þ 2y ¼ 8 are inconsistent, as indicated in Fig. 15-2. Note that if equation (3) is multiplied by 2 we obtain 2x þ 2y ¼ 4, which is obviously inconsistent with (4). Dependent equations are represented by the same line. Thus every point on the line represents a solution and, since there are an infinite number of points, there are an infinite number of simultaneous solutions. For example, (5) x þ y ¼ 1 and (6) 4x þ 4y ¼ 4 are dependent equations as indicated in Fig. 15-3. Note that if (5) is multiplied by 4 the result is (6).

Fig. 15-1

15.2

Fig. 15-2

Fig. 15-3

SYSTEMS OF THREE LINEAR EQUATIONS

A system of three linear equations in three variables is solved by eliminating one unknown from any two of the equations and then eliminating the same unknown from any other pair of equations. Linear equations in three variables represent planes and can result in two or more parallel planes, which are thus inconsistent and have no solution. The three planes can coincide or all three can intersect in a common line and be dependent. The three planes can intersect in a single point, like the ceiling and two walls forming a corner in a room, and are consistent. Linear equations in three variables x, y, and z are of the form ax þ by þ cz ¼ d, where a, b, c, and d are real numbers and not all three of a, b, and c are zero. If we consider three such equations a1 x þ b1 y þ c1 z ¼ d1 a2 x þ b2 y þ c 2 z ¼ d2 a3 x þ b3 y þ c 3 z ¼ d3

CHAP. 15]

SIMULTANEOUS LINEAR EQUATIONS

139

and find a value (x, y, z) that satisfies all three equations, we say we have a simultaneous solution to the system of equations. EXAMPLE 15.1.

Solve the system of equations 2x þ 5y þ 4z ¼ 4, x þ 4y þ 3z ¼ 1, and x 2 3y 2 2z ¼ 5. (1) 2x þ 5y þ 4z ¼ 4 (2) x þ 4y þ 3z ¼ 1 (3) x  3y  2z ¼ 5

First we will eliminate x from (1) and (2) and from (2) and (3). 2x þ 5y þ 4z ¼ 4 2x  8y  6z ¼ 2 (4)

3y  2z ¼ 2

x þ 4y þ 3z ¼ 1 x þ 3y þ 2z ¼ 5 7y þ 5z ¼ 4

(5)

Now we eliminate z from equations (4) and (5). 15y  10z ¼ 10 14y þ 10z ¼ 8 (6) y ¼ 2 We solve (6) and get y ¼ 22. Substituting into (4) or (5), we solve for z. (4)

3(2)  2z ¼ 2 þ6  2z ¼ 2 2z ¼ 4 z¼2

Substituting into (1), (2), or (3), we solve for x. (1)

2x þ 5(2) þ 4(2) ¼ 4 2x  10 þ 8 ¼ 4 2x  2 ¼ 4 2x ¼ 6 x¼3

The solution to the system of equations is (3, 22, 2). We check the solution by substituting the point (3, 22, 2) into equations (1), (2), and (3). (1)

2(3) þ 5(2) þ 4(2) ? 4 6  10 þ 8 ? 4 4¼4

(2)

3 þ 4(2) þ 3(2) ? 1 38 þ6 ?1 1¼1

(3) 3  3(2)  2(2) ? 5 3þ6 4 ? 5 5¼5

Thus (3, 22, 2) checks in each of the given equations and is the answer to the problem.

Solved Problems Solve the following systems. 15.1

(1) 2x  y ¼ 4 (2) x þ y ¼ 5 SOLUTION Add (1) and (2) and obtain 3x ¼ 9, x ¼ 3. Now put x ¼ 3 in (1) or (2) and get y ¼ 2. The solution is x ¼ 3, y ¼ 2 or (3, 2).

140

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

Another method. From (1) obtain y ¼ 2x 2 4 and substitute this value into equation (2) to get x þ 2x 2 4 ¼ 5, 3x ¼ 9, x ¼ 3. Now put x ¼ 3 into (1) or (2) and obtain y ¼ 2. Check: 2x 2 y ¼ 2(3) 2 2 ¼ 4 and x þ y ¼ 3 þ 2 ¼ 5. Graphical solution. The graph of a linear equation is a straight line. Since a straight line is determined by two points, we need plot only two points for each equation. However, to insure accuracy we shall plot three points for each line. For 2x  y ¼ 4:

For x þ y ¼ 5:

x

21

0

1

y

26

24

22

x

21

0

1

y

6

5

4

The simultaneous solution is the point of intersection (3, 2) of the lines (see Fig. 15-4).

Fig. 15-4

15.2

(1) 5x þ 2y ¼ 3 (2) 2x þ 3y ¼ 1 SOLUTION To eliminate y, multiply (1) by 3 and (2) by 2 and subtract the results. 3  (1): 15x þ 6y ¼ 9 2  (2): 4x þ 6y ¼ 2 Subtraction:

11x

¼ 11

or

x ¼ 1:

Now put x ¼ 1 in (1) or (2) and obtain y ¼ 21. The simultaneous solution is (1, 21).

15.3

(1) 2x þ 3y ¼ 3 (2) 6y  6x ¼ 1 SOLUTION Rearranging (2), (1) 2x þ 3y ¼ 3 (2) 6x þ 6y ¼ 1

CHAP. 15]

141

SIMULTANEOUS LINEAR EQUATIONS

To eliminate x, multiply (1) by 3 and add the result with (2) to get 3  (1): (2):

6x þ 9y ¼ 9 6x þ 6y ¼ 1 15y ¼ 10

or

y ¼ 2=3:

Now put y ¼ 2/3 into (1) or (2) and obtain x ¼ 1/2. The solution is (1/2, 2/3).

15.4

(1) 5y ¼ 3  2x (2) 3x ¼ 2y þ 1 SOLUTION Solve by substitution. y¼

From (1), Put this value into (2) and obtain

Then and the solution is

15.5

3  2x : 5

  3  2x 11 þ1 or x¼ : 3x ¼ 2 5 19 3  2x 3  2(11=19) 7 y¼ ¼ ¼ 5 5 19

  11 7 , . 19 19

x2 yþ1 þ ¼2 3 6 x þ 3 2y  1  ¼1 (2) 4 2

(1)

SOLUTION To eliminate fractions, multiply (1) by 6 and (2) by 4 and simplify to obtain (11 ) 2x þ y ¼ 15 (21 ) x  4y ¼ 1 Solving, we find x¼

15.6

59 , 9

y¼

17 9

 and the solution is

(1) x  3y ¼ 2a (2) 2x þ y ¼ 5a SOLUTION To eliminate x, multiply (1) by 2 and subtract (2); then y ¼ a/7. To eliminate y, multiply (2) by 3 and add with (1); then x ¼ 17a/7.   17a a The solution is : , 7 7

15.7

(1) 3u þ 2v ¼ 7r þ s (2) 2u  v ¼ 3s Solve for u and v in terms of r and s.

 59 17 , : 9 9

142

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

SOLUTION To eliminate v, multiply (2) by 2 and add with (1); then 7u ¼ 7r þ 7s or u ¼ r þ s. To eliminate u, multiply (1) by 2, (2) by 23, and add the results; then v ¼ 2r 2 s. The solution is (r þ s, 2r 2 s).

15.8

(1) ax þ by ¼ 2a2  3b2 (2) x þ 2y ¼ 2a  6b SOLUTION Multiply (2) by a and subtract from (1); then by  2ay ¼ 6ab  3b2 , y(b  2a) ¼ 3b(2a  b), 3b(2a  b) 3b(b  2a) ¼ ¼ 3b provided b  2a = 0: (b  2a) b  2a Similarly, we obtain x ¼ 2a provided b 2 2a = 0. Check: (1) a(2a) þ b(3b) ¼ 2a2  3b2 , (2) 2a þ 2(3b) ¼ 2a  6b. Note. If b 2 2a ¼ 0, or b ¼ 2a, the given equations become

and y ¼

(11 ) ax þ 2ay ¼ 10a2 (21 ) x þ 2y ¼ 10a which are dependent since (11) may be obtained by multiplying (21) by a. Thus if b ¼ 2a the system possesses an infinite number of solutions, i.e., any values of x and y which satisfy x þ 2y ¼ 210a.

15.9

The sum of two numbers is 28 and their difference is 12. Find the numbers. SOLUTION Let x and y be the two numbers. Then (1) x þ y ¼ 28 and (2) x 2 y ¼ 12. Add (1) and (2) to obtain 2x ¼ 40, x ¼ 20. Subtract (2) from (1) to obtain 2y ¼ 16, y ¼ 8. Note. Of course this problem may also be solved easily by using one unknown. Let the numbers be n and 28 2 n. Then n 2 (28 2 n) ¼ 12 or n ¼ 20, and 28 2 n ¼ 8.

15.10 If the numerator of a certain fraction is increased by 2 and the denominator is increased by 1, the resulting fraction equals 1/2. If, however, the numerator is increased by 1 and the denominator decreased by 2, the resulting fraction equals 3/5. Find the fraction. SOLUTION Let x ¼ numerator, y ¼ denominator, and x/y ¼ required fraction. Then (1)

xþ2 1 ¼ yþ1 2

or

2x  y ¼ 3,

and

(2)

xþ1 3 ¼ y2 5

or

5x  3y ¼ 11:

Solve (1) and (2) simultaneously and obtain x ¼ 2, y ¼ 7. The required fraction is 2/7.

15.11 Two years ago a man was six times as old as his daughter. In 18 years he will be twice as old as his daughter. Determine their present ages. SOLUTION Let x ¼ father’s present age in years, y ¼ daughter’s present age in years. Equation for condition 2 years ago: Equation for condition 18 years hence: Solve (1) and (2) simultaneously and obtain x ¼ 32, y ¼ 7.

(1) (x  2) ¼ 6(y  2): (2) (x þ 18) ¼ 2(y þ 18):

CHAP. 15]

SIMULTANEOUS LINEAR EQUATIONS

143

15.12 Find the two-digit number satisfying the following two conditions. (1) Four times the units digit is six less than twice the tens digit. (2) The number is nine less than three times the number obtained by reversing the digits. SOLUTION Let t ¼ tens digit, u ¼ units digit. The required number ¼ 10t þ u; reversing digits, the new number ¼ 10u þ t. Then (1) 4u ¼ 2t  6

and

(2) 10t þ u ¼ 3(10u þ t)  9:

Solving (1) and (2) simultaneously, t ¼ 7, u ¼ 2, and the required number is 72.

15.13 Five tables and eight chairs cost $115; three tables and five chairs cost $70. Determine the cost of each table and of each chair. SOLUTION Let x ¼ cost of a table, y ¼ cost of a chair. Then (1) 5x þ 8y ¼ $115

(2) 3x þ 5y ¼ $70:

and

Solve (1) and (2) simultaneously and obtain x ¼ $15, y ¼ $5.

15.14 A merchant sold his entire stock of shirts and ties for $1000, the shirts being priced at 3 for $10 and the ties at $2 each. If he had sold only 1/2 of the shirts and 2/3 of the ties he would have collected $600. How many of each kind did he sell? SOLUTION Let s ¼ number of shirts sold, t ¼ number of ties sold. Then (1)

10 s þ 2t ¼ 1000 3

and

(2)

    10 1 2 s þ 2 t ¼ 600: 3 2 3

Solving (1) and (2) simultaneously, s ¼ 120, t ¼ 300.

15.15 An investor has $1100 income from bonds bearing 4% and 5%. If the amounts at 4% and 5% were interchanged she would earn $50 more per year. Find the total sum invested. SOLUTION Let x ¼ amount invested at 4%, y ¼ amount at 5%. Then (1) 0:04x þ 0:05y ¼ 1100

and

(2) 0:05x þ 0:04y ¼ 1150:

Solving (1) and (2) simultaneously, x ¼ $15 000, y ¼ $10 000, and their sum is $25 000.

15.16 Tank A contains a mixture of 10 gallons water and 5 gallons pure alcohol. Tank B has 12 gallons water and 3 gallons alcohol. How many gallons should be taken from each tank and combined in order to obtain an 8 gallon solution containing 25% alcohol by volume?

144

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

SOLUTION In 8 gal of required mixture are 0.25(8) ¼ 2 gal alcohol. Let x, y ¼ volumes taken from tanks A, B respectively; then (1) x þ y ¼ 8. Fraction of alcohol in tank A ¼

5 1 3 1 ¼ , in tank B ¼ ¼ : 10 þ 5 3 12 þ 3 5

Thus in x gal of A are x/3 gal alcohol, and in y gal of B are y/5 gal alcohol; then (2) x/3 þ y/5 ¼ 2. Solving (1) and (2) simultaneously, x ¼ 3 gal, y ¼ 5 gal. Another method, using only one unknown. Let x ¼ volume taken from tank A, 8 2 x ¼ volume taken from tank B. Then 13 x þ 15 (8  x) ¼ 2 from which x ¼ 3 gal, 8 2 x ¼ 5 gal.

15.17 A given alloy contains 20% copper and 5% tin. How many pounds of copper and of tin must be melted with 100 lb of the given alloy to produce another alloy analyzing 30% copper and 10% tin? All percentages are by weight. SOLUTION Let x, y ¼ number of pounds of copper and tin to be added, respectively. In 100 lb of given alloy are 20 lb copper and 5 lb tin. Then, in the new alloy, fraction of copper ¼ fraction of tin

¼

pounds copper pounds alloy

or

(1) 0:30 ¼

20 þ x 100 þ x þ y

pounds tin pounds alloy

or

(2) 0:10 ¼

5þy : 100 þ x þ y

The simultaneous solution of (1) and (2) is x ¼ 17.5 lb copper, y ¼ 7.5 lb tin.

15.18 Determine the rate of a woman’s rowing in still water and the rate of the river current, if it takes her 2 hours to row 9 miles with the current and 6 hours to return against the current. SOLUTION Let x ¼ rate of rowing in still water, y ¼ rate of current. With current: 2 hr  (x þ y) mi=hr ¼ 9 mi Against current: 6 hr  (x  y) mi=hr ¼ 9 mi

(1) 2x þ 2y ¼ 9: (2) 6x  6y ¼ 9:

or or

Solving (1) and (2) simultaneously, x ¼ 3 mi/hr, y ¼ 3/2 mi/hr.

15.19 Two particles move at different but constant speeds along a circle of circumference 276 ft. Starting at the same instant and from the same place, when they move in opposite directions they pass each other every 6 sec and when they move in the same direction they pass each other every 23 sec. Determine their rates. SOLUTION Let x, y ¼ their respective rates in ft/sec. Opposite directions: Same direction:

6 sec  (x þ y) ft/sec ¼ 276 23 sec  (x 2 y) ft/sec ¼ 276

or or

Solving (1) and (2) simultaneously, x ¼ 29 ft/sec, y ¼ 17 ft/sec.

(1) 6x þ 6y ¼ 276. (2) 23x 2 23y ¼ 276.

CHAP. 15]

SIMULTANEOUS LINEAR EQUATIONS

145

15.20 Fahrenheit temperature ¼ m(Celsius temperature) þ n, or F ¼ mC þ n, where m and n are constants. At one atmosphere pressure, the boiling point of water is 212 8F or 100 8C and the freezing point of water is 32 8F or 0 8C. (a) Find m and n. (b) What Fahrenheit temperature corresponds to 2273 8C, the lowest temperature obtainable? SOLUTION (a) (1) 212 ¼ m(100) þ n and (2) 32 ¼ m(0) þ n. Solving, m ¼ 9/5, n ¼ 32. (b) F ¼ 95 C þ 32 ¼ 95 (273) þ 32 ¼ 491:4 þ 32 ¼ 459 8F, to the nearest degree.

Solve the following systems. 15.21 (1) 2x  y þ z ¼ 3 (2) x þ 3y  2z ¼ 11 (3) 3x  2y þ 4z ¼ 1 SOLUTION To eliminate y between (1) and (2) multiply (1) by 3 and add with (2) to obtain (11 ) 7x þ z ¼ 20: To eliminate y between (2) and (3) multiply (2) by 2, (3) by 3, and add the results to get (21 ) 11x þ 8z ¼ 25: Solving (11) and (21) simultaneously, we find x ¼ 3, z ¼ 21. Substituting these values into any of the given equations, we find y ¼ 2. Thus the solution is (3, 2, 21).

15.22 (1)

x y z þ  ¼ 2, 3 2 4

(2)

x y z 1 þ  ¼ , 4 3 2 6

(3)

x y z 23  þ ¼ : 2 4 3 6

SOLUTION To remove fractions, multiply the equations by 12 and obtain the system (11 ) 4x þ 6y  3z ¼ 24 (21 ) 3x þ 4y  6z ¼ 2 (31 ) 6x  3y þ 4z ¼ 46: To eliminate x between (11 ) and (21 ) multiply (11 ) by 3, (21 ) by 24, and add the results to obtain (12 ) 2y þ 15z ¼ 64: To eliminate x between (21 ) and (31 ) multiply (21 ) by 2 and subtract (31 ) to obtain (22 ) 11y  16z ¼ 42: The simultaneous solution of (12 ) and (22 ) is y ¼ 2, z ¼ 4. Substituting these values of y and z into any of the given equations, we find x ¼ 6. Thus the simultaneous solution of the three given equations is (6, 2, 4).

15.23 (1)

1 2 2   ¼ 0, x y z

(2)

2 3 1 þ þ ¼ 1; x y z

(3)

3 1 3   ¼ 3: x y z

146

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

SOLUTION 1 1 ¼ u, ¼ v, x y so that the given equations may be written

Let

1 ¼w z

(11 ) u  2v  2w ¼ 0 (21 ) 2u þ 3v þ w ¼ 1 (31 ) 3u  v  3w ¼ 3 from which we find u ¼ 22, v ¼ 3, w ¼ 24. 1 1 Thus ¼ 2 or x ¼ 1=2, ¼ 3 or y ¼ 1=3, x y The solution is (1=2, 1=3, 1=4). 1 2 2   ¼ 0, Check: (1) 1=2 1=3 1=4

15.24 (1) 3x þ y  z ¼ 4,

(2)

(2) x þ y þ 4z ¼ 3,

1 ¼ 4 or z ¼ 1=4: z

2 3 1 þ þ ¼ 1, 1=2 1=3 1=4

(3)

3 1 3   ¼ 3: 1=2 1=3 1=4

(3) 9x þ 5y þ 10z ¼ 8.

SOLUTION Subtracting (2) from (1), we obtain (11 ) 2x  5z ¼ 1. Multiplying (2) by 5 and subtracting (3), we obtain (21 ) 4x þ 10z ¼ 7. Now (11 ) and (21 ) are inconsistent since (11 ) multiplied by 22 gives 24x þ10z ¼ 22, thus contradicting (21 ). This indicates that the original system is inconsistent and hence has no simultaneous solution.

15.25 A and B working together can do a given job in 4 days, B and C together can do the job in 3 days, and A and C together can do it in 2.4 days. In how many days can each do the job working alone? SOLUTION Let a, b, c ¼ number of days required by each working alone to do the job, respectively. Then 1/a, 1/b, 1/c ¼ fraction of complete job done by each in 1 day, respectively. Thus (1)

1 1 1 þ ¼ , a b 4

(2)

1 1 1 þ ¼ , b c 3

(3)

1 1 1 þ ¼ . a c 2:4

Solving (1), (2), (3) simultaneously, we find a ¼ 6, b ¼ 12, c ¼ 4 days.

Supplementary Problems 15.26

Solve each of the following pairs of simultaneous equations by the methods indicated. 2x  3y ¼ 7 (a) Solve (1) by addition or subtraction, (2) by substitution. 3x þ y ¼ 5 3x  y ¼ 6 (b) Solve (1) graphically, (2) by addition or subtraction. 2x þ 3y ¼ 7 4x þ 2y ¼ 5 (c) Solve (1) graphically, (2) by addition or subtraction, (3) by substitution. 5x  3y ¼ 2

CHAP. 15]

SIMULTANEOUS LINEAR EQUATIONS

147

15.27

Solve each of the following pairs of simultaneous equations by any method. 2x  5y ¼ 10 2x  3y ¼ 9t (a) (e) 4x þ 3y ¼ 7 4x  y ¼ 8t 2x þ y þ 1 ¼ 0 2y  x ¼ 1 (f ) (b) 3x  2y þ 5 ¼ 0 2x þ y ¼ 8 8 2x y > < þ ¼6 2u  v ¼ 5s 3 5 (c) (g) Find u and v in terms of r and s. 3u þ 2v ¼ 7r  4s > : x  y ¼ 4 6 2 5=x  3=y ¼ 1 (h) 2=x þ 1=y ¼ 7 8 2x  1 y þ 2 > < þ ¼4 ax  by ¼ a2 þ b2 3 4 (i) Find x and y in terms of a and b. (d ) x þ 3 x  y > 2bx  ay ¼ 2b2 þ 3ab  a2 :  ¼3 2 3

15.28

Indicate which of the following systems are (1) consistent, (2) dependent, (3) inconsistent. x þ 3y ¼ 4 3x ¼ 2y þ 3 2x  y ¼ 1 (a) (c) (e) 2x  y ¼ 1 x  2y=3 ¼ 1 2y  x ¼ 1 (x þ 2)=4  (y  2)=12 ¼ 5=4 2x  y ¼ 5 (x þ 3)=4 ¼ (2y  1)=6 (f ) (b) (d ) y ¼ 3x  7 2y ¼ 7 þ 4x 3x  4y ¼ 2

15.29

(a) When the first of two numbers is added to twice the second the result is 21, but when the second number is added to twice the first the result is 18. Find the two numbers. (b) If the numerator and denominator of a certain fraction are both increased by 3, the resulting fraction equals 2/3. If, however, the numerator and denominator are both decreased by 2, the resulting fraction equals 1/2. Determine the fraction. (c) Twice the sum of two numbers exceeds three times their difference by 8, while half the sum is one more than the difference. What are the numbers? (d ) If three times the larger of two numbers is divided by the smaller, the quotient is 6 and the remainder is 6. If five times the smaller is divided by the larger, the quotient is 2 and the remainder is 3. Find the numbers.

15.30

(a) Six years ago Bob was four times as old as Mary. In four years he will be twice as old as Mary. How old are they now? (b) A is eleven times as old as B. In a certain number of years A will be five times as old as B, and five years after that she will be three times as old as B. How old are they now?

15.31

(a) Three times the tens digit of a certain two-digit number is two more than four times the units digit.The difference between the given number and the number obtained by reversing the digits is two less than twice the sum of the digits. Find the number. (b) When a certain two-digit number is divided by the number obtained by reversing the digits, the quotient is 2 and the remainder is 7. If the number is divided by the sum of its digits, the quotient is 7 and the remainder 6. Find the number.

15.32

(a) Two pounds of coffee and 3 lb of butter cost $4.20. A month later the price of coffee advanced 10% and that of butter 20%, making the total cost of a similar order $4.86. Determine the original cost of a pound of each. (b) If 3 gallons of Grade A oil are mixed with 7 gal of Grade B oil the resulting mixture is worth 43 ¢/gal. However, if 3 gal of Grade A oil are mixed with 2 gal of Grade B oil the resulting mixture is worth 46 ¢/gal. Find the price per gallon of each grade. (c) An investor has $116 annual income from bonds bearing 3% and 5% interest. Then he buys 25% more of the 3% bonds and 40% more of the 5% bonds, thereby increasing his annual income by $41. Find his initial investment in each type of bond.

148

15.33

SIMULTANEOUS LINEAR EQUATIONS

[CHAP. 15

(a) Tank A contains 32 gallons of solution which is 25% alcohol by volume. Tank B has 50 gal of solution which is 40% alcohol by volume. What volume should be taken from each tank and combined in order to make up 40 gal of solution containing 30% alcohol by volume? (b) Tank A holds 40 gal of a salt solution containing 80 lb of dissolved salt. Tank B has 120 gal of solution containing 60 lb of dissolved salt. What volume should be taken from each tank and combined in order to make up 30 gal of solution having a salt concentration of 1.5 lb/gal? (c) A given alloy contains 10% zinc and 20% copper. How many pounds of zinc and of copper must be melted with 1000 lb of the given alloy to produce another alloy analyzing 20% zinc and 24% copper? All percentages are by weight. (d ) An alloy weighing 600 lb is composed of 100 lb copper and 50 lb tin. Another alloy weighing 1000 lb is composed of 300 lb copper and 150 lb tin. What weights of copper and tin must be melted with the two given alloys to produce a third alloy analyzing 32% copper and 28% tin. All percentages are by weight.

15.34

(a) Determine the speed of a motorboat in still water and the speed of the river current, if it takes 3 hr to travel a distance of 45 mi upstream and 2 hr to travel 50 mi downstream. (b) When two cars race around a circular mile track starting from the same place and at the same instant, they pass each other every 18 seconds when traveling in opposite directions and every 90 seconds when traveling in the same direction. Find their speeds in mi/hr. (c) A passenger on the front of train A observes that she passes the complete length of train B in 33 seconds when traveling in the same direction as B and in 3 seconds when traveling in the opposite direction. If B is 330 ft long, find the speeds of the two trains.

15.35

Solve each of the following systems of equations. 8 8x y < 2x  y þ 2z ¼ 8 > þ z¼7 > > > (a) x þ 2y  3z ¼ 9

4 2 2 > > >x y z 8 :   ¼1 < x ¼ y  2z 6 4 3 (b) 2y ¼ x þ 3z þ 1 : z ¼ 2y  2x  3

15.36

Indicate which of the following systems are (1) consistent, (2) dependent, (3) inconsistent. 8 8 8 < 2x  y þ z ¼ 1 < x þ y þ 2z ¼ 3

> þ þ ¼5 > >x y z > >

x y z > > > > 3 2 1 > : þ  ¼ 6 x y z

ANSWERS TO SUPPLEMENTARY PROBLEMS 15.26

(a) x ¼ 2, y ¼ 1

(b) x ¼ 1, y ¼ 3

(c) x ¼ 1=2, y ¼ 3=2

15.27

(a) x ¼ 5=2, y ¼ 1

(d ) x ¼ 5, y ¼ 2

(g) u ¼ r  2s, v ¼ 2r þ s

(b) x ¼ 3, y ¼ 2

(e) x ¼ 3t=2, y ¼ 2t

(h) x ¼ 1=2, y ¼ 1=3

(c) x ¼ 6, y ¼ 10

( f ) x ¼ 1, y ¼ 1

(i) x ¼ a þ b, y ¼ a  b

if a2 = 2b2

CHAP. 15]

15.28

149

SIMULTANEOUS LINEAR EQUATIONS

(a) Consistent,

(c) Dependent

(e) Consistent,

(b) Inconsistent,

(d ) Inconsistent,

( f ) Dependent.

15.29

(a) 5, 8

(b) 7/12

(c) 7, 3

15.30

(a) Mary 11 yr, Bob 26 yr

15.31

(a) 64

15.32

(a) Coffee 90 ¢/lb, butter 80 ¢/lb

(d) 16, 7

(b) A is 22 yr, B is 2 yr

(b) 83 (b) Grade A 50 ¢/gal, Grade B 40 ¢/gal

(c) $1200 at 3%, $1600 at 5% 15.33

(a) 26 2/3 gal from A

(b) 20 gal from A

13 1/3 gal from B

10 gal from B

(c) 150 lb zinc 100 lb copper

15.34

(a) Boat 20 mi/hr, river 5 mi/hr

15.35

(a) x ¼ 21, y ¼ 2, z ¼ 22

(c) x ¼ 6, y ¼ 4, z ¼ 23

(b) x ¼ 0, y ¼ 2, z ¼ 1

(d) x ¼ 1/2, y ¼ 21/3, z ¼ 1/6

15.36

(a) Dependent

15.37

4, 2, 3

15.38

371

(b) Inconsistent

(b) 120 mi/hr, 80 mi/hr

(c) Consistent

(d) 400 lb copper 500 lb tin (c) 60 ft/sec, 50 ft/sec

CHAPTER 16

Quadratic Equations in One Variable 16.1

QUADRATIC EQUATIONS

A quadratic equation in the variable x has the form ax2 þ bx þ c ¼ 0 where a, b, and c are constants and a = 0. Thus x2  6x þ 5 ¼ 0, 2x2 þ x  6 ¼ 0, x2 þ 3x ¼ 0, and 3x2  5 ¼ 0 are quadratic equations in one variable. The last two equations may be divided by 2 and 4 respectively to obtain x2 þ 12x  3 ¼ 0 and x2  53 ¼ 0 where the coefficient of x2 in each case is 1. An incomplete quadratic equation is one which either b ¼ 0 or c ¼ 0, e.g., 4x2  5 ¼ 0, 7x2  2x ¼ 0, and 3x2 ¼ 0. To solve a quadratic equation ax2 þ bx þ c ¼ 0 is to find values of x which satisfy the equation. These values of x are called zeros or roots of the equation. For example, x2  5x þ 6 ¼ 0 is satisfied by x ¼ 2 and x ¼ 3. Then x ¼ 2 and x ¼ 3 are zeros or roots of the equation. 16.2 A.

METHODS OF SOLVING QUADRATIC EQUATIONS Solution by square root EXAMPLES 16.1. (a) x  4 ¼ 0 2

Solve each quadratic equation for x.

(b) 2x2  21 ¼ 0

(c) x2 þ 9 ¼ 0

(a) x2  4 ¼ 0. Then x2 ¼ 4, x ¼ +2, and the roots are x ¼ 2, 2. pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi (b) 2x2  21 ¼ 0. Then x2 ¼ 21=2 and the roots are x ¼ + 21=2 ¼ +12 42: p ffiffiffiffiffiffi ffi (c) x2 þ 9 ¼ 0. Then x2 ¼ 9 and the roots are x ¼ + 9 ¼ +3i:

B.

Solution by factoring EXAMPLES 16.2.

Solve each quadratic equation for x.

(a) 7x  5x ¼ 0

(b) x2  5x þ 6 ¼ 0

2

(c) 3x2 þ 2x  5 ¼ 0

(d ) x2  4x þ 4 ¼ 0

(a) 7x2  5x ¼ 0 may be written as x(7x  5) ¼ 0. Since the product of the two factors is zero, we set each factor equal to 0 and solve the resulting linear equations. x ¼ 0 or 7x  5 ¼ 0. So x ¼ 0 and x ¼ 5=7 are the roots of the equation.

150

CHAP. 16]

QUADRATIC EQUATIONS IN ONE VARIABLE

151

(b) x2  5x þ 6 ¼ 0 may be written as (x  3)(x  2) ¼ 0. Since the product is equal to 0, we set each factor equal to 0 and solve the resulting linear equations. x  3 ¼ 0 or x  2 ¼ 0. So x ¼ 3 and x ¼ 2 are the roots of the equation. (c) 3x2 þ 2x  5 ¼ 0 may be written as (3x þ 5)(x  1) ¼ 0. Thus, 3x þ 5 ¼ 0 or x  1 ¼ 0 and the roots of the equation are x ¼ 5=3 and x ¼ 1. (d ) x2  4x þ 4 ¼ 0 may be written as (x  2)(x  2) ¼ 0. Thus, x  2 ¼ 0 and the equation has a double root x ¼ 2.

C.

Solution by completing the square EXAMPLE 16.3.

Solve x2  6x  2 ¼ 0.

Write the unknowns on one side and the constant term on the other; then x2  6x ¼ 2:

Add 9 to both sides, thus making the left-hand side a perfect square; then x2  6x þ 9 ¼ 2 þ 9

or

(x  3)2 ¼ 11:

pffiffiffiffiffi pffiffiffiffiffi Hence x  3 ¼ + 11 and the required roots are x ¼ 3 + 11. Note. In the method of completing the square (1) the coefficient of the x2 term must be 1 and (2) the number added to both sides is the square of half the coefficient of x. EXAMPLE 16.4.

Solve 3x2  5x þ 1 ¼ 0:

Dividing by 3,

 2 1 5 25 ¼  to both sides, Adding 2 3 36

x2 

5x 1 ¼ : 3 3

5 25 1 25 13 x  xþ ¼ þ ¼ , 3 36 3 36 36 pffiffiffiffiffi 13 5 x ¼+ and 6 6 2

D.



 5 2 13 ¼ , x 6 36 pffiffiffiffiffi 13 5 x¼ + : 6 6

Solution by quadratic formula. The solutions of the quadratic equation ax2 þ bx þ c ¼ 0 are given by the formula x¼

b +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac 2a

where b2  4ac is called the discriminant of the quadratic equation. For a derivation of the quadratic formula see Problem 16.5. EXAMPLE 16.5.

Solve 3x2  5x þ 1 ¼ 0: Here a ¼ 3, b ¼ 5, c ¼ 1 so that

x¼

(5) +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi (5)2  4(3)(1) 5 + 13 ¼ 2(3) 6

as in Example 16.4.

152

QUADRATIC EQUATIONS IN ONE VARIABLE

[CHAP. 16

EXAMPLE 16.6. Solve 4x2  6x þ 3 ¼ 0: Here a ¼ 4, b ¼ 6, and c ¼ 3. x¼

(6) +

pffiffiffi 3+i 3 x¼ 4

E.

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi (6)2  4(4)(3) 6 + 12 6 + 2i 3 2(3 + i 3) ¼ ¼ ¼ 8 2(4) 8 8

Graphical solution The real roots or zeros of ax2 þ bx þ c ¼ 0 are the values of x corresponding to y ¼ 0 on the graph of the parabola y ¼ ax2 þ bx þ c: Thus the solutions are the abscissas of the points where the parabola intersects the x axis. If the graph does not intersect the x axis the roots are not real.

16.3

SUM AND PRODUCT OF THE ROOTS

The sum S and the product P of the roots of the quadratic equation ax2 þ bx þ c ¼ 0 are given by S ¼ b=a and P ¼ c=a. Thus in 2x2 þ 7x  6 ¼ 0 we have a ¼ 2, b ¼ 7, c ¼ 6 so that S ¼ 7=2 and P ¼ 6=2 ¼ 3: It follows that a quadratic equation whose roots are r1 , r2 is given by x2  Sx þ P ¼ 0 where S ¼ r1 þ r2 and P ¼ r1 r2 . Thus the quadratic equation whose roots are x ¼ 2 and x ¼ 5 is x2  (2  5)x þ 2(5) ¼ 0 or x2 þ 3x  10 ¼ 0:

16.4

NATURE OF THE ROOTS

The nature of the roots of the quadratic equation ax2 þ bx þ c ¼ 0 is determined by the discriminant b2  4ac: When the roots involve the imaginary unit i, we say the roots are not real. Assuming a, b, c are real numbers then (1) if b2  4ac . 0, the roots are real and unequal, (2) if b2  4ac ¼ 0, the roots are real and equal, (3) if b2  4ac , 0, the roots are not real. Assuming a, b, c are rational numbers then (1) (2) (3) (4)

if if if if

b2  4ac is a perfect square = 0, the roots are real, rational, and unequal, b2  4ac ¼ 0, the roots are real, rational, and equal, b2  4ac . 0 but not a perfect square, the roots are real, irrational, and unequal, b2  4ac , 0, the roots are not real.

Thus 2x2 þ 7x  6 ¼ 0, with discriminant b2  4ac ¼ 72  4(2)(6) ¼ 97, has roots which are real, irrational and unequal.

16.5

RADICAL EQUATIONS

A radical equation havingpone orffi more unknowns under a radical. pffiffiffiffiffiffiffiffiffiffiffi ispan ffiffiffiffiffiffiffiffiffiffi ffiffiffi equation p ffiffiffi Thus x þ 3  x ¼ 1 and 3 y ¼ y  4 are radical equations. To solve a radical equation, isolate one of the radical terms on one side of the equation and transpose all other terms to the other side. If both members of the equation are then raised to a power equal to the index of the isolated radical, the radical will be removed. This process is continued until radicals are no longer present.

CHAP. 16]

153

QUADRATIC EQUATIONS IN ONE VARIABLE

EXAMPLE 16.7.

Solve

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi x þ 3  x ¼ 1: pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Transposing, x þ 3 ¼ x þ 1. pffiffiffi pffiffiffi Squaring, x þ 3 ¼ x þ 2 x þ 1 or x ¼ 1. pffiffiffi Finally, squaring both sides of x ¼ 1 gives x ¼ 1. pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Check: 1 þ 3  1 ? 1, 2  1 ¼ 1:

It is very important to check the values obtained, as this method often introduces extraneous roots which are to be rejected. 16.6

QUADRATIC-TYPE EQUATIONS

An equation of quadratic type has the form az2n þ bzn þ c ¼ 0 where a = 0, b, c, and n = 0 are constants, and where z depends on x. Upon letting zn ¼ u this equation becomes au2 þ bu þ c ¼ 0, which may be solved for u. These values of u may be used to obtain z from which it may be possible to obtain x. EXAMPLE 16.8.

Solve x4  3x2  10 ¼ 0: Let u ¼ x2 and substituting Factoring Solving for u Substituting x2 ¼ u Solving for x

EXAMPLE 16.9.

u2  3u  10 ¼ 0 (u  5)(u  2) ¼ 0 u ¼ 5 or u ¼ 2 x2 ¼ 5 p orffiffiffi x2 ¼ 2 pffiffiffi x ¼ + 5 or x ¼ +i 2

Solve (2x  1)2 þ 7(2x  1) þ 12 ¼ 0. Let u ¼ 2x  1 and substituting Factoring Solving for u Substituting 2x  1 ¼ u Solving for x

u2 þ 7u þ 12 ¼ 0 (u þ 4)(u þ 3) ¼ 0 u ¼ 4 or u ¼ 3 2x  1 ¼ 4 or 2x  1 ¼ 3 x ¼ 3=2 or x ¼ 1

Solved Problems 16.1

Solve (a) x2  16 ¼ 0.

Then x2 ¼ 16, x ¼ +4.

(b) 4t  9 ¼ 0. Then 4t2 ¼ 9, t2 ¼ 9=4, t ¼ +3=2. pffiffiffi pffiffiffiffiffiffiffiffi (c) 3  x2 ¼ 2x2 þ 1. Then 3x2 ¼ 2, x2 ¼ 2=3, x ¼ + 2=3 ¼ +13 6. pffiffiffiffiffiffiffiffiffiffiffi (d) 4x2 þ 9 ¼ 0. Then x2 ¼ 9=4, x ¼ + 9=4 ¼ +32i: 2x2  1 17 ¼xþ3þ . Then 2x2  1 ¼ (x þ 3)(x  3) þ 17, 2x2  1 ¼ x2  9 þ 17, (e) x3 x3 x2 ¼ 9, and x ¼ +3. 2

Check. If x ¼ 3 is substituted into the original equation, we have division by zero which is not allowed. Hence x ¼ 3 is not a solution. 2(3)2  1 17 17 17 ?3þ3þ or ¼ and x ¼ 3 is a solution. If x ¼ 3, 3  3 3  3 6 6 16.2

Solve by factoring (a) (b) (c) (d)

x2 þ 5x  6 ¼ 0, (x þ 6)(x  1) ¼ 0, x ¼ 6, 1: t2 ¼ 4t, t2  4t ¼ 0, t(t  4) ¼ 0, t ¼ 0, 4. x2 þ 3x ¼ 28, x2 þ 3x  28 ¼ 0, (x þ 7)(x  4) ¼ 0, 2 2x2  5x þ 2 ¼ 0, (2x  1)(x  2) ¼ 0, 5x  2x ¼ 2,

x ¼ 7, 4: x ¼ 1=2, 2:

154

QUADRATIC EQUATIONS IN ONE VARIABLE

(e)

1 1 5 þ ¼ . t1 t4 4

[CHAP. 16

Multiplying by 4(t  1)(t  4),

4(t  4) þ 4(t  1) ¼ 5(t  1)(t  4), 5t2  33t þ 40 ¼ 0, (t  5)(5t  8) ¼ 0, t ¼ 5, 8=5: y 3p (f) ¼ , 6y2  5py  6p2 ¼ 0, (3y þ 2p)(2y  3p) ¼ 0, y ¼ 2p=3, 3p=2: 2p 6y  5p 16.3

What term must be added to each of the following expressions in order to make it a perfect square trinomial?

2

2 (a) x2  2x. Add 12(coefficient of x) ¼ 12(2) ¼ 1. Check: x2  2x þ 1 ¼ (x  1)2 . (b) x2 þ 4x.

Add

2(coefficient

2

2 of x) ¼ 12(4) ¼ 4.

5 (c) u þ u. 4

 2 1 5 25 Add ¼ . 2 4 64

(d ) x4 þ px2 .

Add

2

16.4

1

Check: x2 þ 4x þ 4 ¼ (x þ 2)2 .

  5 25 5 2 Check: u þ u þ ¼ uþ . 4 64 8 2

1 2 2 2(p) ¼ p =4.

Check: x4 þ px2 þ p2 =4 ¼ (x2 þ p=2)2 .

Solve by completing the square. (a) x2  6x þ 8 ¼ 0. Hence

Then x2  6x ¼ 8,

x  3 ¼ +1,

Check: For x ¼ 4,

x2  6x þ 9 ¼ 8 þ 9,

(x  3)2 ¼ 1:

x ¼ 3 + 1, and the roots are x ¼ 4 and x ¼ 2.

4  6(4) þ 8 ? 0,

For x ¼ 2, 22  6(2) þ 8 ? 0, 0 ¼ 0:  2  2   3 3 3 2 25 2 2 2 Then t þ 3t ¼ 4, t þ 3t þ ¼4þ , tþ ¼ : (b) t ¼ 4  3t. 2 2 2 4 3 5 3 5 t¼ + , and the roots are t ¼ 1, 4: Hence t þ ¼ + , 2 2 2 2  2  2   8 5 8 4 5 4 4 2 1 2 2 2 (c) 3x þ 8x þ 5 ¼ 0: x þ xþ Then x þ x ¼  , ¼ þ , xþ ¼ : 3 3 3 3 3 3 3 9 4 1 4 1 Hence x þ ¼ + , x¼ + , and the roots are x ¼ 1, 5=3: 3 3 3 3 2

0 ¼ 0.

(d ) x2 þ 4x þ 1 ¼ 0. Then x2 þ 4x ¼ 1, x2 þ 4x þ 4 ¼ 3, (x þ 2)2 ¼ 3: pffiffiffi pffiffiffi Hence x þ 2 ¼ + 3, and roots are x ¼ 2 + 3. pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi Check: For x ¼ 2 þ 3, (2 þ 3)2 þ 4(2 þ 3) þ 1 ¼ (4  4 3 þ 3)  8 þ 4 3 þ 1 ¼ 0. pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi For x ¼ 2  3, (2  3)2 þ 4(2  3) þ 1 ¼ (4 þ 4 3 þ 3)  8  4 3 þ 1 ¼ 0.  2  2   6x 3 3 3 2 16 2 2 2 Then 5x  6x ¼ 5, x  þ ¼ 1 þ , x ¼ : (e) 5x  6x þ 5 ¼ 0. 5 5 5 5 25 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 4 Hence x  3=5 ¼ + 16=25, and the roots are x ¼ + i. 5 5 16.5

Solve the equation ax2 þ bx þ c ¼ 0, a=0, by the method of completing the square. SOLUTION Dividing both sides by a, Adding

x2 þ

b c xþ ¼0 a a

  2 1 b b2 ¼ 2 to both sides, 2 a 4a

x2 þ

or

x2 þ

b c x¼ : a a

b b2 c b2 b2  4ac : xþ 2 ¼ þ 2 ¼ a a 4a 4a2 4a

CHAP. 16]

  b 2 b2  4ac Then x þ ¼ , 2a 4a2

16.6

155

QUADRATIC EQUATIONS IN ONE VARIABLE

xþ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b b2  4ac ¼+ , 2a 2a

and

x¼

b +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac : 2a

Solve by the quadratic formula. (a) x2  3x þ 2 ¼ 0.

Here a ¼ 1, b ¼ 3, c ¼ 2. Then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b + b2  4ac (3) + (3)2  4(1)(2) 3 + 1 ¼ ¼ x¼ 2 2a 2(1)

(b) 4t2 þ 12t þ 9 ¼ 0: Here a ¼ 4, b ¼ 12, c ¼ 9. Then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 12 + (12)  4(4)(9) 12 + 0 3 ¼ and t¼ ¼ 8 2 2(4)

t¼

or

x ¼ 1, 2:

3 is a double root: 2

(c) 9x2 þ 18x  17 ¼ 0: Here a ¼ 9, b ¼ 18, c ¼ 17. Then pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffi 18 + (18)2  4(9)(17) 18 + 936 18 + 6 26 3 + 26 ¼ ¼ ¼ : x¼ 2(9) 18 18 3 (d) 6u(2  u) ¼ 7.

Then 6u2  12u þ 7 ¼ 0 and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffi pffiffiffi 6 (12) + (12)2  4(6)(7) 12 + 24 12 + 2 6i i: ¼ ¼1+ ¼ u¼ 6 12 12 2(6)

16.7

Solve graphically: (a) 2x2 þ 3x  5 ¼ 0,

(b) 4x2  12x þ 9 ¼ 0,

(c) 4x2  4x þ 5 ¼ 0:

SOLUTION (a) y ¼ 2x2 þ 3x  5

x

23

22

21

0

1

2

y

4

23

26

25

0

9

The graph of y ¼ 2x2 þ 3x  5 indicates that when y ¼ 0, x ¼ 1, and 2:5. Thus the roots of 2x2 þ 3x  5 ¼ 0 are x ¼ 1, 2:5 (see Fig. 16-1(a)).

(b) y ¼ 4x2  12x þ 9

x

21

0

1

2

3

4

y

25

9

1

1

9

25

The graph of y ¼ 4x2  12x þ 9 is tangent to the x axis at x ¼ 1:5, i.e., when y ¼ 0, x ¼ 1:5. Thus 4x2  12x þ 9 ¼ 0 has the equal roots x ¼ 1:5 (see Fig. 16-1(b)).

(c) y ¼ 4x2  4x þ 5

x

22

21

0

1

2

3

y

29

13

5

5

13

29

The graph of y ¼ 4x2  4x þ 5 does not intersect the x axis, i.e., there is no real value of x for which y ¼ 0. Hence the roots of 4x2  4x þ 5 ¼ 0 are not real (see Fig. 16-1(c)). (By the quadratic formula the roots are x ¼ 12 + i:)

156

QUADRATIC EQUATIONS IN ONE VARIABLE

[CHAP. 16

Fig. 16-1

16.8

Prove that the sum S and product P of the roots of the quadratic equation ax2 þ bx þ c ¼ 0 are S ¼ b=a and P ¼ c=a. SOLUTION By the quadratic formula the roots are b þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac 2a

and

b 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2  4ac : 2a

The sum of the roots is S¼

2b b ¼ : 2a a

The product of the roots is P¼

16.9

b þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi! b2  4ac b  b2  4ac (b)2  (b2  4ac) c ¼ : ¼ 2a 2a 4a2 a

Without solving, find the sum S and product P of the roots. b c Here a ¼ 1, b ¼ 7, c ¼ 6; then S ¼  ¼ 7, P ¼ ¼ 6: a a 6 3 : 2x2 þ 6x  3 ¼ 0: Here a ¼ 2, b ¼ 6, c ¼ 3; then S ¼  ¼ 3, P ¼ 2 2 1 5 x þ 3x2 þ 5 ¼ 0: Write as 3x2 þ x þ 5 ¼ 0: Then S ¼  , P ¼ : 3 3 5 2 3x  5x ¼ 0: Here a ¼ 3, b ¼ 5, c ¼ 0; then S ¼ , P ¼ 0. 3 3 2 2x þ 3 ¼ 0: Here a ¼ 2, b ¼ 0, c ¼ 3; then S ¼ 0, P ¼ . 2 m 2 þ n2 mn 2 2 2 ¼ 1. mnx þ (m þ n )x þ mn ¼ 0. , P¼ Then S ¼  mn mn 0:01 1 4 40 0:3x2  0:01x þ 4 ¼ 0. ¼ , P¼ ¼ : Then S ¼  0:3 30 0:3 3

(a) x2  7x þ 6 ¼ 0: (b) (c) (d ) (e) (f) (g)

CHAP. 16]

QUADRATIC EQUATIONS IN ONE VARIABLE

157

16.10 Find the discriminant b2  4ac of each of the following equations and thus determine the character of their roots. (a) (b) (c) (d) (e) (f )

x2  8x þ 12 ¼ 0. b2  4ac ¼ (8)2  4(1)(12) ¼ 16; the roots are real, rational, unequal. 2 3y þ 2y  4 ¼ 0. b2  4ac ¼ 52; the roots are real, irrational, unequal. b2  4ac ¼ 31; the roots are conjugate not real numbers. 2x2  x þ 4 ¼ 0: 2 4z  12z þ 9 ¼ 0. b2  4ac ¼ 0; the roots are real, rational, equal. b2  4ac ¼ 12; the roots are conjugate not real numbers. 2x  4x2 ¼ 1 or 4x2  2x þ 1 ¼ 0. pffiffiffi 2 pffiffiffi pffiffiffi 2x  4 3x þ 4 2 ¼ 0: Here the coefficients are real but not rational numbers. b2  4ac ¼ 16; the roots are real and unequal.

16.11 Find a quadratic equation with integer coefficients having the given pair of roots. (S ¼ sum of roots, P ¼ product of roots.) (a) 1, 2 Method 1. S ¼ 1 þ 2 ¼ 3, P ¼ 2; hence x2  3x þ 2 ¼ 0: Method 2. (x  1) and (x  2) must be factors of the quadratic expression. Then (x  1)(x  2) ¼ 0 or x2  3x þ 2 ¼ 0: (b) 3, 2 Method 1. S ¼ 1, P ¼ 6; hence x2 þ x  6 ¼ 0: Method 2. ½x  (3) and (x  2) are factors of the quadratic expression. Then (x þ 3)(x  2) ¼ 0 or x2 þ x  6 ¼ 0: 4 3 11 4 11 4 S¼ ,P¼ ; hence x2  x ¼0 or 15x2  11x  12 ¼ 0: (c) ,  . 3 5 15 5 15 5 pffiffiffi pffiffiffi (d) 2 þ 2, 2  2 pffiffiffi pffiffiffi 2 Method 1. S ¼ 4, P ¼ p(2 ffiffiffi þ 2)(2  2)p¼ffiffiffi 2; hence x  4x þ 2 ¼ 0. Method 2. ½x  (2 þ 2) and ½x are expression. pffiffi ffi (2  2) p ffiffiffi factors of the quadratic pffiffiffi pffiffiffi Then ½x  (2 þ 2)½x  (2  2) ¼ ½(x  2)  2½(x  2) þ 2 ¼ 0, 2 2 (x  2) pffiffiffi 2 ¼ 0 or x pffiffiffi 4x þ 2 ¼ 0: Method 3. Since x ¼ 2 + 2, x  2 ¼ + 2. Squaring, (x  2)2 ¼ 2 or x2  4x þ 2 ¼ 0: (e) 3 þ 2i, 3 2i Method 1. S ¼ 6, P ¼ (3 þ 2i)(3  2i) ¼ 13; hence x2 þ 6x þ 13 ¼ 0: Method 2. ½x  (3 þ 2iÞ and ½x  (3  2i) are factors of the quadratic expression. Then ½(x þ 3)  2i½(x þ 3) þ 2i ¼ 0, (x þ 3)2 þ 4 ¼ 0 or x2 þ 6x þ 13 ¼ 0. 16.12 In each quadratic equation find the value of the constant p subject to the given condition. (a) 2x2  px þ 4 ¼ 0 has one root equal to 3. Since x ¼ 3 is a root, it must satisfy the given equation. Then 2(3)2  p(3) þ 4 ¼ 0 and p ¼ 22=3: (b) (p þ 2)x2 þ 5x þ 2p ¼ 0 has the product of its roots equal to 2/3. The product of the roots is 2p ; pþ2

then

2p 2 ¼ pþ2 3

and

p ¼ 1:

(c) 2px2 þ px þ 2x ¼ x2 þ 7p þ 1 has the sum of its roots equal to 4=3: Write the equation in the form (2p  1)x2 þ (p þ 2)x  (7p þ 1) ¼ 0: Then the sum of the roots is 

pþ2 4 ¼ 2p  1 3

and

p ¼ 2:

(d) 3x2 þ ( p þ 1) þ 24 ¼ 0 has one root equal to twice the other. Let the roots be r, 2r.

158

QUADRATIC EQUATIONS IN ONE VARIABLE

[CHAP. 16

Product of the roots is r(2r) ¼ 8; then r 2 ¼ 4 and r ¼ +2. Sum of the roots is 3r ¼ ( p þ 1)=3. Substitute r ¼ 2 and r ¼ 22 into this equation and obtain p ¼ 19 and p ¼ 17 respectively. (e) 2x2  12x þ p þ 2 ¼ 0 has the difference between its roots equal to 2. Let the roots be r, s; then (1) r  s ¼ 2: The sum of the roots is 6; then (2) r þ s ¼ 6. The simultaneous solution of (1) and (2) is r ¼ 4, s ¼ 2. Now put x ¼ 2 or x ¼ 4 into the given equation and obtain p ¼ 14. 16.13 Find the roots of each quadratic equation subject to the given conditions. (a) (2k þ 2)x2 þ (4  4k)x þ k  2 ¼ 0 has roots which are reciprocals of each other. Let r and 1/r be the roots, their product being 1. k2 ¼ 1, from which k ¼ 4. Product of roots is 2k þ 2 Put k ¼ 4 into the given equation; then 3x2  10x þ 3 ¼ 0 and the roots are 1/3, 3. (b) kx2  (1 þ k)x þ 3k þ 2 ¼ 0 has the sum of its roots equal to twice the product of its roots. Sum of roots ¼ 2(product of roots); then   1þk 3k þ 2 3 ¼2 and k¼ : k k 5 Put k ¼ 3=5 into the given equation; then 3x2 þ 2x  1 ¼ 0 and the roots are 1, 1=3. (c) (x þ k)2 ¼ 2  3k has equal roots. Write the equation as x2 þ 2kx þ (k2 þ 3k  2) ¼ 0 where a ¼ 1, b ¼ 2k, c ¼ k2 þ 3k  2. The roots are equal if the discriminant (b2  4ac) ¼ 0: Then from b2  4ac ¼ (2k)2  4(1)(k2 þ 3k  2) ¼ 0 we get k ¼ 2=3. Put k ¼ 2=3 into the given equation and solve to obtain the double root 2=3: 16.14 Solve pffiffiffiffiffiffiffiffiffiffiffiffiffi (a) 2x þ 1 ¼ 3:pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Squaring both sides, 2x þ 1 ¼ 9 and x ¼ 4. ffi Check. 2(4) þ 1 ? 3, 3 ¼ 3: pffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 5 þ 2x ¼ x þ 1. Squaring bothffi sides, 5 þ 2x ¼ x2 þ 2x þ 1, x2 ¼ 4 and x ¼ +2. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Check. For x ¼ 2, 5pþffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2(2) ? 2 þ 1 or 3 ¼ 3. pffiffiffi pffiffiffi For x ¼ 2, 5 þ 2(2) ?  2 þ 1 or 1 ¼ 1 which is not true since 1 ¼ 1: Thus x ¼ 2 is the only solution; x ¼ 2 is an extraneous root. pffiffiffiffiffiffiffiffiffiffiffiffiffi þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6 ¼ 0 and (c) 3x  5 ¼ x  1: Squaring, 3xffi  5 ¼ x2  2x þ 1, x2  5xp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi x ¼ 3, 2. Check. For x ¼ 3, 3(3)  5 ? 3  1 or 2 ¼ 2. For x ¼ 2, 3(2)  5 ? 2  1 or 1 ¼ 1. Thus both x ¼ 3 and x ¼ 2 are solutions of the given equation. p p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3 2 x2  x þ 6 ¼ 2, x2  x þ 6 ¼ 8, x2  x  2 ¼ 0 and x ¼ 2, 1. (d ) x  x þ 6  2 ¼ 0. Then ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 3 2 Check. For x ¼ 2, p 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 þ 6  2 ? 0 or ffi 2  2 ¼ 0. For x ¼ 1, 3 (1)2  (1) þ 6  2 ? 0 or 2  2 ¼ 0. 16.15 Solve pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (a) 2x þ 1  x ¼ 1. Rearranging, (1) 2x þ p 1¼ ffiffiffi x þ 1. pffiffiffi Squaring both sides of (1), 2x þ 1 ¼ x þ 2 x þ 1 or (2) x ¼ 2 x. Squaring (2), x2 ¼ 4x; then x(xffi  p 4)ffiffiffi¼ 0 and x ¼ 0, 4. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi Check. For x ¼ 0, 2(0) þ 1  0 ?1, 1 ¼ 1. For x ¼ 4, 2(4) þ 1  4 ?1, 1 ¼ 1. pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 4x  1 þ 2x þ 3 ¼ 1. Rearranging, 2x þ 3. pffiffiffiffiffiffiffiffiffiffiffiffiffi(1) 4x  1 ¼ 1 pffiffiffiffiffiffiffiffiffiffiffiffiffi Squaring (1), 4x  1 ¼ 1  2 2x þ 3 þ 2x þ 3 or (2) 2 2x þ 3 ¼ 5  2x. Squaring (2), 4(2x þ 3) ¼ 25  20x þ 4x2 , 4x2  28x þ 13 ¼ 0 and x ¼ 1=2, 13=2.

CHAP. 16]

QUADRATIC EQUATIONS IN ONE VARIABLE

159

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi For x ¼ 1=2, p4(1=2)  1 þffi 2(1=2) þ 3 ? 1 ffi or 3 ¼ 1 which is not true. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi For x ¼ 13=2, 4(13=2)  1 þ 2(13=2) þ 3 ? 1 or 9 ¼ 1 which is not true. Hence x ¼ 1=2 and x ¼ 13=2 are extraneous roots; the equation has no solution. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi (c) x þ 16  x ¼ 2. Squaring,pffiffiffi x þ 16 pffiffiffi x ¼ 4 or (1) x þ 16 ¼ x þ 4. Squaring (1), x þ 16 ¼ x þ 8 x þ 16, 8 x ¼ 0, and x ¼ 0 is a solution. Check.

16.16 Solve pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi (a) x2 þ 6x ¼ x þ 2x. pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi Squaring, x2 þ 6x ¼ x2pþffiffiffiffiffi2x 2x þ 2x, 2x 2x ¼ 4x, x( 2x  2) ¼ 0. pffiffiffiffiffi Then x ¼ 0; and from 2x  2 ¼ 0, 2x ¼ 2, 2x ¼ 4, x ¼ 2. Both x ¼ 0 and x ¼ 2 satisfy the given equation. pffiffiffi pffiffiffi pffiffiffi 2 (b) x  pffiffiffi ¼ 1. Multiply by x and obtain (1) x  2 ¼ x. x Squaring (1), x2  4x þ 4 ¼ x, x2  5x þ 4 ¼ 0, (x  1)(x  4) ¼ 0, and x ¼ 1, 4. Only x ¼ 4 satisfies the given equation; x ¼ 1 is extraneous. 16.17 Solve the equation x2  6x 

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2  6x  3 ¼ 5.

SOLUTION

pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi Let x2  6x ¼ u; then u  u  3 ¼ 5 or (1) u  3 ¼ u  5. Squaring (1), u  3 ¼ u2  10u þ 25, u2  11u þ 28 ¼ 0, and u ¼ 7, 4. Since only u ¼ 7 satisfies (1), substitute u ¼ 7 in x2  6x ¼ u and obtain x2  6x  7 ¼ 0,

(x  7)(x þ 1) ¼ 0,

and

x ¼ 7, 1:

Both x ¼ 7 and x ¼ 1 satisfy the original equation and are thus solutions. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Note. If we write the given equation as x2  6x  3 ¼ x2  6x  5 and square both sides, the resulting fourth degree equation would be difficult to solve.

16.18 Solve the equation 4x 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ : 2 x  8x þ 32 5 SOLUTION Squaring, 16  8x þ x2 9 ¼ ; x2  8x þ 32 25 then 25(16  8x þ x2 ) ¼ 9(x2  8x þ 32), x2  8x þ 7 ¼ 0, and x ¼ 7, 1. The only solution is x ¼ 1; reject x ¼ 7, an extraneous solution.

16.19 Solve Let x2 ¼ u; then u2  10u þ 9 ¼ 0 and u ¼ 1, 9. (a) x4  10x2 þ 9 ¼ 0. 2 For u ¼ 1, x ¼ 1 and x ¼ +1; for u ¼ 9, x2 ¼ 9 and x ¼ +3. The four solutions are x ¼ +1, +3; each satisfies the given equation. (b) 2x4 þ x2  1 ¼ 0. Let x2 ¼ u; then 2u2 þ u  1 ¼ 0 and u ¼ 12 , 1. pffiffiffi 1 2 1 If u ¼ 2 , x ¼ 2 and x ¼ + 12 p 2;ffiffiffiif u ¼ 1, x2 ¼ 1 and x ¼ +i. 1 The four solutions are x ¼ +2 2, +i.

160

QUADRATIC EQUATIONS IN ONE VARIABLE

[CHAP. 16

pffiffiffi pffiffiffi p ffiffiffi Let 4 x ¼ u; then u2  u  2 ¼ 0 and u ¼ 2, 1. x  4 x  2 ¼ 0. pffiffiffi pffiffiffi If u ¼ 2, 4 x ¼ 2 and x ¼ 24 ¼ 16. Since 4 x is positive, it cannot equal 21. Hence x ¼ 16 is the only solution of the given equation.     1 2 1 (d ) 2 x þ þ 5 ¼ 0. 7 x þ x x 1 Let x þ ¼ u; then 2u2  7u þ 5 ¼ 0 and u ¼ 5=2, 1. x 5 1 5 For u ¼ , x þ ¼ , 2x2  5x þ 2 ¼ 0 and x ¼ 2, 12. 2 x 2 pffiffiffi 1 For u ¼ 1, x þ ¼ 1, x2  x þ 1 ¼ 0 and x ¼ 12 + 12 3i. x pffiffiffi The four solutions are x ¼ 2, 12 , 12 + 12 3i. (c)

(e) 9(x þ 2)4 þ 17(x þ 2)2  2 ¼ 0. Let (x þ 2)2 ¼ u; then 9u2 þ 17u  2 ¼ 0 and u ¼ 1=9, 2. If (x þ 2)2 ¼ 1=9, (x þ 2)2 ¼ 9, (x þ 2) ¼ +3 and x ¼ 1, 5. pffiffiffi pffiffiffi If (x þ 2)2 ¼ 2, (x þ 2)2 ¼  12 , (x þ 2) ¼ + 12 2i and x ¼ 2 + 12 2i. pffiffiffi The four solutions are x ¼ 1, 5, 2 + 12 2i. 16.20 Find the values of x which satisfy each of the following equations.  4  2 x x (a) 16  25 þ 9 ¼ 0: xþ1 xþ1  2 x ¼ u; then 16u2  25u þ 9 ¼ 0 and u ¼ 1, 9=16. Let xþ1  2 x x ¼ +1. ¼1 or If u ¼ 1, xþ1 xþ1 x ¼1 The equation xþ1 x ¼ 1 has no solution; the equation xþ1 has solution x ¼ 1=2. If u ¼ 9=16,  2 x 9 ¼ xþ1 16

or

x 3 ¼ + so that x ¼ 3, 3=7: xþ1 4

The required solutions are x = 1=2, 3=7, 3. (b) (x þ 3x þ 2)2  8(x2 þ 3x) ¼ 4. Let x2 þ 3x ¼ u; then (u þ 2)2  8u ¼ 4 and u ¼ 0, 4. 2 If u ¼ 0, x þ 3x ¼ 0 and x ¼ 0, 3; if u ¼ 4, x2 þ 3x ¼ 4 and x ¼ 4, 1. The solutions are x ¼ 4, 3, 0, 1. 2

16.21 One positive number exceeds three times another positive number by 5. The product of the two numbers is 68. Find the numbers. SOLUTION Let x ¼ smaller number; then 3x þ 5 ¼ larger number. Then x(3x þ 5) ¼ 68, 3x2 þ 5x  68 ¼ 0, (3x þ 17)(x  4) ¼ 0, and x ¼ 4, 17=3. We exclude 17=3 since the problem states that the numbers are positive. The required numbers are x ¼ 4 and 3x þ 5 ¼ 17.

CHAP. 16]

QUADRATIC EQUATIONS IN ONE VARIABLE

161

16.22 When three times a certain number is added to twice its reciprocal, the result is 5. Find the number. SOLUTION Let x ¼ the required number and 1=x ¼ its reciprocal. Then 3x þ 2(1=x) ¼ 5, 3x2  5x þ 2 ¼ 0, (3x  2)(x  1) ¼ 0, and x ¼ 1, 2=3. Check. For x ¼ 1, 3(1) þ 2(1=1) ¼ 5; for x ¼ 2=3, 3(2=3) þ 2(3=2) ¼ 5.

16.23 Determine the dimensions of a rectangle having perimeter 50 feet and area 150 square feet. SOLUTION Sum of four sides ¼ 50 ft; hence, sum of two adjacent sides ¼ 25 ft (see Fig. 16-2). Let x and 25  x be the lengths of two adjacent sides. The area is x(25  x) ¼ 150; then x2  25x þ 150 ¼ 0, (x  10)(x  15) ¼ 0, and x ¼ 10, 15. Then 25  x ¼ 15, 10; and the rectangle has dimensions 10 ft by 15 ft.

16.24 The hypotenuse of a right triangle is 34 inches. Find the lengths of the two legs if one leg is 14 inches longer than the other. SOLUTION Let x and x þ 14 be the lengths of the legs (see Fig. 16-3). Then x2 þ (x þ 14)2 ¼ (34)2 , x2 þ 14x  480 ¼ 0, (x þ 30)(x  16) ¼ 0, and x ¼ 30, 16. Since x ¼ 30 has no physical significance, we have x ¼ 16 in. and x þ 14 ¼ 30 in.

Fig. 16-2

Fig. 16-3

Fig. 16-4

16.25 A picture frame of uniform width has outer dimensions 12 in. by 15 in. Find the width of the frame (a) if 88 square inches of picture show, (b) if 100 square inches of picture show. SOLUTION Let x ¼ width of frame; then the dimensions of the picture are (15  2x), (12  2x) (see Fig. 16-4). (a) Area of picture ¼ (15 2 2x)(12 2 2x) ¼ 88; then 2x2  27x þ 46 ¼ 0, (x  2)(2x  23) ¼ 0, and x ¼ 2, 1112. Clearly, the width cannot be 1112 in. Hence the width of the frame is 2 in. Check. The area of the picture is (15 2 4)(12 2 4) ¼ 88 in.2, as given. (b) Here (15 2 2x)(12 22x) ¼ 100, 2x 2 2 27x þ 40 ¼ 0 and, by the quadratic formula, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi b + b2  4ac 27 + 409 ¼ or x ¼ 11:8, 1:7 (approximately): x¼ 2a 4 Reject x ¼ 11.8 in., which cannot be the width. The required width is 1.7 in.

16.26 A pilot flies a distance of 600 miles. He could fly the same distance in 30 minutes less time by increasing his average speed by 40 mi/hr. Find his actual average speed.

162

QUADRATIC EQUATIONS IN ONE VARIABLE

[CHAP. 16

SOLUTION Let x ¼ actual average speed in mi/hr. Time in hours ¼

distance in mi : speed in mi=hr

Time to fly 600 mi at x mi/hr 2 time to fly 600 mi at (x þ 40) mi/hr ¼ 12 hr. 600 600 1  ¼ . x x þ 40 2

Then

Solving, the required speed is x ¼ 200 mi/hr.

16.27 A retailer bought a number of shirts for $180 and sold all but 6 at a profit of $2 per shirt. With the total amount received she could buy 30 more shirts than before. Find the cost per shirt. SOLUTION Let x ¼ cost per shirt in dollars; 180/x ¼ number of shirts bought.  Then

   180 180  6 (x þ 2) ¼ x þ 30 : x x

Solving, x ¼ $3 per shirt.

16.28 A and B working together can do a job in 10 days. It takes A 5 days longer than B to do the job when each works alone. How many days would it take each of them, working alone, to do the job? SOLUTION Let n, n 2 5 ¼ number of days required by A and B respectively, working alone, to do the job. In 1 day A does 1/n of job and B does l/(n 2 5) of job. Thus in 10 days they do together   1 1 10 þ ¼ 1 complete job: n n5 Then 10(2n 2 5) ¼ n(n 2 5), n 2 2 25n þ 50 ¼ 0, and n¼

25 +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 625  200 ¼ 22:8, 2:2: 2

Rejecting n ¼ 2.2, the required solution is n ¼ 22.8 days, n 2 5 ¼ 17.8 days.

16.29 A ball projected vertically upward with initial speed v0 ft/sec is at time t sec at a distance s ft from the point of projection as given by the formula s ¼ v0t 2 16t 2. If the ball is given an initial upward speed of 128 ft/sec, at what times would it be 100 ft above the point of projection? SOLUTION s ¼ v0 t  16t2 ,

100 ¼ 128t  16t2 ,

4t2  32t þ 25 ¼ 0, and t ¼

pffiffiffiffiffiffiffiffi 32 + 624 ¼ 7:12, 0:88: 8

At t ¼ 0.88 sec, s ¼ 100 ft and the ball is rising; at t ¼ 7.12 sec, s ¼ 100 ft and the ball is falling. This is seen from the graph of s plotted against t (see Fig. 16-5).

CHAP. 16]

163

QUADRATIC EQUATIONS IN ONE VARIABLE

Fig. 16-5

Supplementary Problems 16.30

Solve each equation. (a) x2  40 ¼ 9

(d )

x 4 ¼ 16 x

(f )

1  2x x2 ¼ 3x 3x  1

(e)

y2 y2 ¼ þ2 3 6

(g)

1 1 1  ¼ 2x  1 2x þ 1 4

(b) 2x2  400 ¼ 0 (c) x2 þ 36 ¼ 9  2x2 16.31

16.32

16.33

16.34

16.35

16.36

(h) x 

2x 5 ¼ 1 xþ1 xþ1

Solve each equation by factoring. (a) x2  7x ¼ 12

(d ) 2x2 þ 2 ¼ 5x

(b) x2 þ x ¼ 6

(e) 9x2 ¼ 9x  2

(c) x2 ¼ 5x þ 24

( f ) 4x  5x2 ¼ 12

(g)

x 4a ¼ 2a x þ 2a

(i)

2x  1 x þ 2 10 þ ¼ x þ 2 2x  1 3

(h)

1 1 1  ¼ 4x 2þx 4

( j)

2c  3y y 2  ¼ yc 2y  c 3

Solve each equation by completing the square. (a) x2 þ 4x  5 ¼ 0

(c) 2x2 ¼ x þ 1

(e) 4x2 ¼ 12x  7

(g) 2x2 þ 3a2 ¼ 7ax

(b) x(x  3) ¼ 4

(d ) 3x2  2 ¼ 5x

( f ) 6y2 ¼ 19y  15

(h) 12x  9x2 ¼ 5

Solve each equation by the quadratic formula. (a) x2  5x ¼ 6

(d ) 16x2  8x þ 1 ¼ 0

(b) x2  6 ¼ x

(e) x(5x  4) ¼ 2

(c) 3x2  2x ¼ 8

( f ) 9x2 þ 6x ¼ 4

(g)

5x2  2p2 p ¼ x 3

(h)

2x þ 3 3x  2 ¼ 4x  1 3x þ 2

Solve each equation graphically. (a) 2x2 þ x  3 ¼ 0

(c) x2  2x ¼ 2

(e) 6x2  7x  5 ¼ 0

(b) 4x2  8x þ 4 ¼ 0

(d ) 2x2 þ 2 ¼ 3x

( f ) 2x2 þ 8x þ 3 ¼ 0

Without solving, find the sum S and product P of the roots of each equation. (a) 2x2 þ 3x þ 1 ¼ 0

(d ) 2x2 þ 6x  5 ¼ 0

(g) 2x2 þ 5kx þ 3k2 ¼ 0

(b) x  x2 ¼ 2

(e) 3x2  4 ¼ 0

(c) 2x(x þ 3) ¼ 1

( f ) 4x2 þ 3x ¼ 0

(h) 0:2x2  0:1x þ 0:03 ¼ 0 pffiffiffi pffiffiffi (i) 2x2  3x þ 1 ¼ 0

Find the discriminant b2  4ac and thus determine the character of the roots. (a) 2x2  7x þ 4 ¼ 0

(c) 3x  x2 ¼ 4

(b) 3x2 ¼ 5x  2

(d ) x(4x þ 3) ¼ 5

(e) 2x2 ¼ 5 þ 3x pffiffiffi ( f ) 4x 3 ¼ 4x2 þ 3

(g) 1 þ 2x ¼ 2x2 ¼ 0 (h) 3x þ 25=3x ¼ 10

164

QUADRATIC EQUATIONS IN ONE VARIABLE

16.37

Find a quadratic equation with integer coefficients (if possible) having the given roots. pffiffiffi pffiffiffi pffiffiffi pffiffiffi (a) 2, 3 (d ) 2, 5 (g) 1 þ i, 1 i ( j) 3  2, 3 þ 2 pffiffiffi pffiffiffi (b) 3, 0 (e) 1=3, 1=2 (h) 2  6, 2 þ 6 (k) a þ bi, a 2 bi a, b integers pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 mþ n m n (c) 8, 4 ( f ) 2 þ 3, 2  3 (i) 2 þ i, 2  i (l) , m, n integers 2 2 2 2

16.38

In each quadratic equation, evaluate the constant p subject to the given condition.

[CHAP. 16

(a) px2  x þ 5  3p ¼ 0 has one root equal to 2. (b) (2p þ 1)x2 þ px þ p ¼ 4( px þ 2) has the sum of its roots equal to the product of its roots. (c) 3x2 þ p(x  2) þ 1 ¼ 0 has roots which are reciprocals. (d ) 4x2  8x þ 2p  1 ¼ 0 has one root equal to three times the other. (e) 4x2  20x þ p2  4 ¼ 0 has one root equal to two more than the other. ( f ) x2 ¼ 5x  3p þ 3 has the difference between its roots equal to 11. 16.39

Find the roots of each equation subject to the given condition. (a) 2px2  4px þ 5p ¼ 3x2 þ x  8 has the product of its roots equal to twice their sum. (b) x2  3(x  p)  2 ¼ 0 has one root equal to 3 less than twice the other root. (c) p(x2 þ 3x  9) ¼ x  x2 has one root equal to the negative of the other. (d ) (m þ 3)x2 þ 2m(x þ 1) þ 3 ¼ 0 has one root equal to half the reciprocal of the other. (e) (2m þ 1)x2  4mx ¼ 1  3m has equal roots.

16.40

16.41

Solve each equation. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) x2  x þ 2 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 2x  2 ¼ x  1 pffiffiffiffiffiffiffiffiffiffiffiffiffi (c) 4x þ 1 ¼ 3  3x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d ) 2  3 x2 þ 2x ¼ 0

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi x2  2x þ 1 ¼ 2  x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ( j) 2x  10 þ x þ 9 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (k) 2x þ 8 þ 2x þ 5 ¼ 8x þ 25 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (l) 3 2x  1 ¼ 6 x þ 1 (i)

Solve the equation. (a) x4  13x2 þ 36 ¼ 0 (b) x4  3x2  10 ¼ 0 (c) 4x4  17x2 þ 4 ¼ 0 (d ) x4=3  5x2=3 þ 4 ¼ 0

16.42

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2x þ 7 ¼ x þ 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( f ) 2x2  7  x ¼ 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (g) 2 þ x  4 þ 10  3x ¼ 0 pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 (h) 2 x  4x  3 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi 4x  3 (e)

(e) (x2  6x)2  2(x2  6x) ¼ 35 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( f ) x2 þ x ¼ 7 x2 þ x þ 2  12     1 2 7 1 ¼2  (g) x þ xþ x 2 x

(h)

pffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffi xþ2 4 xþ2¼6

(i) x3  7x3=2  8 ¼ 0 ( j)

x2 þ 2 8x þ 2 ¼6 x x þ2

(a) The sum of the squares of two numbers is 34, the first number being one less than twice the second number. Determine the numbers. (b) The sum of the squares of three consecutive integers is 110. Find the numbers. (c) The difference between two positive numbers is 3, and the sum of their reciprocals is 1/2. Determine the numbers. (d ) A number exceeds twice its square root by 3. Find the number.

16.43

(a) The length of a rectangle is three times its width. If the width is diminished by 1 ft and the length increased by 3 ft, the area will be 72 ft2. Find the dimensions of the original rectangle. (b) A piece of wire 60 in. long is bent into the form of a right triangle having hypotenuse 25 in. Find the other two sides of the triangle.

CHAP. 16]

165

QUADRATIC EQUATIONS IN ONE VARIABLE

(c) A picture 8 in. by 12 in. is placed in a frame which has uniform width. If the area of the frame equals the area of the picture, find the width of the frame. (d ) A rectangular piece of tin 9 in. by 12 in. is to be made into an open box with base area 60 in.2 by cutting out equal squares from the four corners and then bending up the edges. Find to the nearest tenth of an inch the length of the side of the square cut from each corner. 16.44

(a) The tens digit of a certain two-digit number is twice the units digit. If the number is multiplied by the sum of its digits, the product is 63. Find the number. (b) Find a number consisting of two digits such that the tens digit exceeds the units digit by 3 and the number is 4 less than the sum of the squares of its digits.

16.45

(a) Two men start at the same time from the same place and travel along roads that are at right angles to each other. One man travels 4 mi/hr faster than the other, and at the end of 2 hours they are 40 miles apart. Determine their rates of travel. (b) By increasing her average speed by 10 mi/hr a motorist could save 36 minutes in traveling a distance of 120 miles. Find her actual average speed. (c) A woman travels 36 miles down a river and back in 8 hours. If the rate of her boat in still water is 12 mi/hr, what is the rate of the river current?

16.46

(a) A merchant purchased a number of coats, each at the same price, for a total of $720. He sold them at $40 each, thus realizing a profit equal to his cost of 8 coats. How many did he buy? (b) A grocer bought a number of cans of corn for $14.40. Later the price increased 2 cents a can and as a result she received 24 fewer cans for the same amount of money. How many cans were in his first purchase and what was the cost per can?

16.47

(a) It takes B 6 hours longer than A to assemble a machine. Together they can do it in 4 hr. How long would it take each working alone to do the job? (b) Pipe A can fill a given tank in 4 hr. If pipe B works alone, it takes 3 hr longer to fill the tank than if pipes A and B act together. How long will it take pipe B working alone?

16.48

A ball projected vertically upward is distance s ft from the point of projection after t seconds, where s ¼ 64t 2 16t 2. (a) At what times will the ball be 40 ft above the ground? (b) Will the ball ever be 80 ft above the ground? (c) What is the maximum height reached?

ANSWERS TO SUPPLEMENTARY PROBLEMS 16.30

16.31

16.32

(a) x ¼ +7

pffiffiffi (b) x ¼ +10 2

pffiffiffi (e) y ¼ +2 3

(g) x ¼ +3=2

(d ) x ¼ +8

( f ) x ¼ +1

(h) x ¼ +2

(a) 3, 4

(c) 8, 23

(e) 1/3, 2/3

(g) 2a, 4a

(i) 1, 7

(b) 2, 3

(d ) 2, 1/2

( f ) 2, 6=5

(h) 2, 8

( j) 2c/5, 4c/5

(a) 1, 5 (b) 4, 1

16.33

(c) x ¼ +3i

(c) 1, 21/2 (d ) 2, 1=3

(a) 6, 21

(c) 2, 24/3

(b) 3, 22

(d ) 1/4, 1/4

(e)

(e)

pffiffiffi 3+ 2 2

pffiffiffiffiffi 2 + 14 5

( f ) 3/2, 5/3 (g) 3a, a/2

(f)

pffiffiffi 1 + i 3 5

(h)

2 i + 3 3

(g)

2p 3p ; 3 5

(h)

pffiffiffiffiffi 6 + 42 3

166

16.34

QUADRATIC EQUATIONS IN ONE VARIABLE

(a) x ¼ 3=2 and x ¼ 1 (see Fig. 16-6). (b) Double root of x ¼ 1 (see Fig. 16-7). (c) Real zeros between 1 and 0 and between 2 and 3 (see Fig. 16-8). (d ) No real zeros (see Fig. 16-9).

Fig. 16-6

Fig. 16-8

Fig. 16-7

Fig. 16-9

(e) Real zeros between 21 and 0 and between 1 and 2 (see Fig. 16-10). ( f ) Real zeros between 24 and 23 and between 21 and 0 (see Fig. 16-11).

[CHAP. 16

CHAP. 16]

16.35

16.36

16.37

16.38

(a) S ¼ 3=2, P ¼ 1=2

(d ) S ¼ 3, P ¼ 5=2

(g) S ¼ 5k=2, P ¼ 3k2 =2

(b) S ¼ 1, P ¼ 2

(e) S ¼ 0, P ¼ 4=3

(c) S ¼ 3, P ¼ 1=2

( f ) S ¼ 3=4, P ¼ 0

(h) S ¼ 0:5, P ¼ 0:15 pffiffiffi pffiffiffi (i) S ¼ 12 6, P ¼ 12 2

(a) 17; real, irrational, unequal

(e) 49; real, rational, unequal

(b) 1; real, rational, unequal

( f ) 0; real, equal

(c) 27; not real

(g) 24; not real

(d ) 89; real, irrational, unequal

(h) 0; real, rational, equal

(a) x2 þ x  6 ¼ 0

(e) 6x2  x  1 ¼ 0

(b) x2 þ 3x ¼ 0

( f ) x2  4x þ 1 ¼ 0

pffiffiffi ( j) not possible (x2  2 3x þ 1 ¼ 0)

(c) x2  4x  32 ¼ 0

(g) x2 þ 2x þ 2 ¼ 0

(k) x2  2ax þ a2 þ b2 ¼ 0

(d ) x2 þ 7x þ 10 ¼ 0

(h) x2 þ 4x  2 ¼ 0

(l) 4x2  4mx þ m2  n ¼ 0

(a) p ¼ 23

(b) p ¼ 24

(c) p ¼ 21

(i) 4x2  16x þ 25 ¼ 0

(d ) p ¼ 2

(e) p ¼ +5

Fig. 16-10

16.39

(a) 3, 6

16.41

(d) 1/2 + i/2 if m ¼ 1/2, the roots are 1/2, 1/2.

(a) 2, 21

(c) 4/9

(e) 9, 1

(g) +2

(i) 3/2

(k) 22

(b) 1, 3

(d ) 24, 2

( f ) 8, 22

(h) 1

( j ) no solution

(l) 5/4

(a) +2, +3 pffiffiffi pffiffiffi (b) + 5, + i 2 (c) +2, +1=2

16.42

( f ) p ¼ 27

Fig. 16-11

(c) +3/2

(b) 1, 2

(e) If m ¼ 21, the roots are 2, 2; 16.40

167

QUADRATIC EQUATIONS IN ONE VARIABLE

(d ) +1, +1=8

pffiffiffiffiffi pffiffiffi (g) 2 + 3, 1=4 + i 15=4

(e) 7, 5, +1

(h) 79

pffiffiffiffiffi ( f ) 1, 2, (1 + 93)=2

(a) 5, 3 or 227/5, 211/5

(i) 4

(b) 5, 6, 7 or 27, 26, 25

(c) 3, 6

(d) 9

pffiffiffi ( j) 1 + i, 2 + 2

168

QUADRATIC EQUATIONS IN ONE VARIABLE

16.43

(a) 5, 15 ft

(b) 15, 20 in.

(c) 2 in.

16.44

(a) 21

16.45

(a) 12, 16 mi/hr

16.46

(a) 24

16.47

(a) A, 6 hr; B, 12 hr

16.48

(a) 0.78 and 3.22 seconds after projection

(d ) 1.3 in.

(b) 85 (b) 40 mi/hr

(c) 6 mi/hr

(b) 144, 10¢ (b) 5.3 hr approx. (b) No

(c) 64 ft

[CHAP. 16

CHAPTER 17

Conic Sections 17.1

GENERAL QUADRATIC EQUATIONS

The general quadratic equation in the two variables x and y has the form ax2 þ bxy þ cy2 þ dx þ ey þ f ¼ 0

(1)

where a, b, c, d, e, f are given constants and a, b, c are not all zero. Thus 3x2 þ 5xy ¼ 2, x2  xy þ y2 þ 2x þ 3y ¼ 0, y2 ¼ 4x, xy ¼ 4 are quadratic equations in x and y. The graph of equation (1), if a, b, c, d, e, f are real, depends on the value of b2  4ac. (1) (2) (3)

If b2  4ac , 0, the graph is in general an ellipse. However, if b ¼ 0 and a ¼ c the graph may be a circle, a point, or non-existent. The point and non-existent situations are called the degenerate cases. If b2  4ac ¼ 0, the graph is a parabola, two parallel or coincident lines, or non-existent. The parallel or coincident lines and non-existent situations are called the degenerate cases. If b2  4ac . 0, the graph is a hyperbola or two intersecting lines. The two intersecting lines situation is called the degenerate case.

These graphs are the intersections of a plane and a right circular cone, and for this reason are called conic sections. EXAMPLES 17.1.

Identify the type of conic section described by each equation.

(a) x2 þ xy ¼ 6 (b) x2 þ 5xy  4y2 ¼ 10

(c) 2x2  y2 ¼ 7 (d ) 3x2 þ 2y2 ¼ 14

(e) 3x2 þ 3y2  4x þ 3y þ 10 ¼ 0 ( f ) y2 þ 4x þ 3y þ 4 ¼ 0

(a) a ¼ 1, b ¼ 1, c ¼ 0 b2  4ac ¼ 1  4 , 0 So the figure is an ellipse or a degenerate case. (b) a ¼ 1, b ¼ 5, c ¼ 4 b2  4ac ¼ 25 þ 16 . 0 So the figure is a hyperbola or a degenerate case. (c) a ¼ 2, b ¼ 0, c ¼ 1 b2  4ac ¼ 0 þ 8 . 0 So the figure is a hyperbola or a degenerate case. (d ) a ¼ 3, b ¼ 0, c ¼ 2 b2  4ac ¼ 0  24 , 0 So the figure is an ellipse or a degenerate case. (e) a ¼ 3, b ¼ 0, c ¼ 3 b2  4ac ¼ 0  36 , 0 So the figure is a circle or a degenerate case since a ¼ c and b ¼ 0. ( f ) a ¼ 0, b ¼ 0, c ¼ 1 b2  4ac ¼ 0  0 ¼ 0 So the figure is a parabola or a degenerate case.

169

170

CONIC SECTIONS

17.2

[CHAP. 17

CONIC SECTIONS

Each conic section is the locus (set) of all points in a plane meeting a given set of conditions. The set of points can be described by an equation. When the locus is positioned at the origin, the figure is called a central conic section. A general equation used to describe a conic section is called the standard equation, which may have more than one form for a conic section. The conic sections are the circle, parabola, ellipse, and hyperbola. We will consider only conic sections in which b ¼ 0, thus having the general quadratic equation Ax2 þ Cy2 þ Dx þ Ey þ F ¼ 0. Trigonometry is needed to discuss fully the general quadratic equations in which b = 0. 17.3

CIRCLES

A circle is the locus of all points in a plane which are at a fixed distance from a fixed point in the plane. The fixed point is the center of the circle and the fixed distance is the radius of the circle. When the center of the circle is the origin, (0, 0), and the radius is r, the standard form of the equation of a circle is x2 þ y2 ¼ r 2 . If the center of the circle is the point (h, k) and the radius is r, the standard form of the equation of a circle is (x  h)2 þ (y  k)2 ¼ r 2 . If r 2 ¼ 0, we have the degenerate case of a single point which is sometimes called the point circle. If r 2 , 0, we have the non-existent degenerate case, which is sometimes called the imaginary circle, since the radius would have to be an imaginary number. The graph of the circle (x  2)2 þ (y þ 3)2 ¼ 9 has its center at (2, 3) and a radius of 3 (see Fig. 17-1).

Fig. 17-1

EXAMPLES 17.2.

For each circle state the center and radius.

(a) x2 þ y2 ¼ 5

(b) x2 þ y2 ¼ 28 (c) (x þ 2)2 þ (y  4)2 ¼ 81 pffiffiffi (a) C(0, 0), r ¼ 5 pffiffiffiffiffi pffiffiffipffiffiffi pffiffiffi (b) C(0, 0), r ¼ 28 ¼ 4 7 ¼ 2 7 (c) (x þ 2)2 þ (y  4)2 ¼ 81 so (x  (2))2 þ (y  4)2 ¼ 92 EXAMPLES 17.3.

Write the equation of each circle in standard form.

(a) x þ y  8x þ 12y  48 ¼ 0 2

2

C(2, 4), r ¼ 9

(b)

x2 þ y2  4x þ 6y þ 100 ¼ 0

CHAP. 17]

(a)

(b)

CONIC SECTIONS

x2 þ y2  8x þ 12y  48 ¼ 0 (x2  8x) þ (y2 þ 12y) ¼ 48 (x2  8x þ 16) þ (y2 þ 12y þ 36) ¼ 48 þ 16 þ 36 (x  4)2 þ (y þ 6)2 ¼ 100

rearrange terms complete the square for x and y standard form (1)

x2 þ y2  4x þ 6y þ 100 ¼ 0 (x2  4x) þ (y2 þ 6y) ¼ 100 (x2  4x þ 4) þ (y2 þ 6y þ 9) ¼ 100 þ 4 þ 9 (x  2)2 þ (y þ 3) ¼ 87

rearrange terms complete the square for x and y standard form (2)

171

Note: In (1) r 2 ¼ 100, so we have a circle, but in (2) r 2 ¼ 87 so we have the degenerate case. EXAMPLE 17.4. Write the equation of the circle going through the points P(2, 1), Q(3, 0), and R(1, 4). By substituting the points P, Q, and R into the general form of a circle, x2 þ y2 þ Dx þ Ey þ F ¼ 0, we get a system of three linear equations. for P(2, 1) for Q(3, 0) for R(1, 4)

22 þ (1)2 þ 2D  E þ F ¼ 0 (3)2 þ 02  3D þ 0E þ F ¼ 0 12 þ 42 þ D þ 4E þ F ¼ 0

then (1) then (2) then (3)

2D  E þ F ¼ 5 3D þ F ¼ 9 D þ 4E þ F ¼ 17

Eliminating F from (1) and (2) and from (1) and (3), we get (4) 5D  E ¼ 4

and

(5) D  5E ¼ 12

Solving (4) and (5) we get D ¼ 1/3 and E ¼ 7=3, and by substituting D and E in (1) we get F ¼ 8. The equation of the circle is x2 þ y2 þ 1=3x  7=3y  8 ¼ 0 or 3x2 þ 3y2 þ x  7y  24 ¼ 0:

17.4

PARABOLAS

A parabola is the locus of all points in a plane equidistant from a fixed line, the directrix, and a fixed point, the focus. Central parabolas have their vertex at the origin, focus on one axis, and directrix parallel to the other axis. We denote the distance from the focus to the vertex by jpj. The distance from the directrix to the vertex is also jpj. The equations of the central parabolas are (1) and (2) below. (1)

y2 ¼ 4pix

and

(2) x2 ¼ 4py

In (1) the focus is on the x axis and the directrix is parallel to the other axis. If p is positive, the curve opens to the right and if p is negative, the curve opens to the left (see Fig. 17-2). In (2) the focus is on the y axis and the directrix is parallel to the x axis. If p is positive, the curve opens up and if p is negative, the curve opens down (see Fig. 17-3). The line through the vertex and the focus is the axis of the parabola and the graph is symmetric with respect to this line.

Fig. 17-2

172

CONIC SECTIONS

[CHAP. 17

Fig. 17-3

The parabolas with vertex at the point (h, k) and with the axis and the directrix parallel to the x axis and the y axis have the standard forms listed in (3) and (4) below. (3) (y  k)2 ¼ 4p(x  h)

and

(4) (x  h)2 ¼ 4p(y  k)

In (3) the focus is F(h þ p, k), the directrix is x ¼ h  p, and the axis is y ¼ k (see Fig. 17- 4). However, in (4) the focus is F(h, k þ p), the directrix is y ¼ k  p, and the axis is x ¼ h (see Fig. 17-5).

Fig. 17-4

Fig. 17-5

CHAP. 17]

EXAMPLES 17.5. (a)

y2 ¼ 8x

173

CONIC SECTIONS

Determine the vertex, focus, directrix, and axis for each parabola. (b)

x2 ¼ 6y

(c)

(y  3)2 ¼ 5(x þ 7)

(d )

(x  1)2 ¼ 4(y þ 4)

y2 ¼ 8x: vertex (h, k) ¼ (0, 0), 4p ¼ 8, so p ¼ 2, focus ( p, 0) ¼ (2, 0), and directrix is x ¼ p, so x ¼ (2) ¼ 2, axis is y ¼ 0 (b) x2 ¼ 6y: vertex (h, k) ¼ (0, 0), 4p ¼ 6, so p ¼ 3/2, focus (0, p) ¼ (0, 3/2), and directrix is y ¼ p, so y ¼ 3=2, axis is y ¼ 0 (c) (y  3)2 ¼ 5(x þ 7): vertex (h, k) ¼ (7, 3), 4p ¼ 5, so p ¼ 5/4, focus (h þ p, k) ¼ (27 þ 5/4, 3) ¼ (223/4, 3), and directrix is x ¼ h  p, so x ¼ 7  5=4 ¼ 33=4, axis y ¼ k, so y ¼ 3 (d ) (x  1)2 ¼ 4(y þ 4): vertex (h, k) ¼ (1, 4), 4p ¼ 4, so p ¼ 1, focus (h, k þ p) ¼ (1, 4 þ (1)) ¼ (1, 5), and directrix is y ¼ k  p, so y ¼ 4  (1) ¼ 3, axis is x ¼ h, so x ¼ 1 (a)

EXAMPLES 17.6.

Write the equation of the parabola with the given characteristics. (b)

focus (3, 5) and directrix y ¼ 3

(a)

vertex (4, 6) and focus (4, 8)

(a)

Since the vertex (4, 6) and the focus (4, 8) lie on the line x ¼ 4 (see Fig. 17-4), we have a parabola of the form (x  h)2 ¼ 4p(y  k): Since the vertex is (4, 6), we have h ¼ 4 and k ¼ 6. The focus is (h, k þ p), so k þ p ¼ 8 and 6 þ p ¼ 8, so p ¼ 2. The equation of the parabola is (x  4)2 ¼ 8(y  6):

(b)

Since the directrix is y ¼ 3 (see Fig. 17-5), the parabola has the form (x  h)2 ¼ 4p(y  k): The focus (3, 5) is 2 units above the directrix y ¼ 3, so p . 0. The distance from the focus to the directrix is 2jpj, so 2p ¼ 2 and p ¼ 1. The focus is (h, p þ k), so h ¼ 3 and k þ p ¼ 5. Since p ¼ 1, k ¼ 4. The equation of the parabola is (x  3)2 ¼ 4(y  4):

EXAMPLES 17.7. Write the equation of each parabola in standard form. (b) y2 þ 3x  6y ¼ 0 (a) x2  4x  12y  32 ¼ 0 (a)

(b)

17.5

x2  4x  12y  32 ¼ 0 x2  4x ¼ 12y þ 32 x2  4x þ 4 ¼ 12y þ 32 þ 4 (x  2)2 ¼ 12y þ 36 (x  2)2 ¼ 12(y þ 3)

reorganize terms complete the square for x factor right-hand side of equation standard form

y2 þ 3x  6y ¼ 0 y2  6y ¼ 3x y2  6y þ 9 ¼ 3x þ 9 (y  3)2 ¼ 3(x  3)

reorganize terms complete the square for y standard form

ELLIPSES

An ellipse is the locus of all points in a plane such that the sum of the distances from two fixed points, the foci, to any point on the locus is a constant. Central ellipses have their center at the origin, vertices and foci lie on one axis, and the covertices lie on the other axis. We will denote the distance from a vertex to the center by a, the distance from a covertex to the center by b, and the distance from a focus to the center by c. For an ellipse, the values a, b, and c are related by a2 ¼ b2 þ c2 and a . b. We call the line segment between the vertices the major axis and the line segment between the covertices the minor axis. The standard forms for the central ellipses are:

(1)

x2 y2 þ ¼1 a2 b2

and

(2)

y2 x2 þ ¼1 a2 b2

174

CONIC SECTIONS

[CHAP. 17

The larger denominator is always a2 for an ellipse. If the numerator for a2 is x2 , then the major axis lies on the x axis. In (1) the vertices have coordinates V(a, 0) and V 0 (a, 0), the foci have coordinates F(c, 0) and F 0 (c, 0), and the covertices have coordinates B(0, b) and B0 (0, b) (see Fig. 17-6). If the numerator for a2 is y2 , then the major axis lies on the y axis. In (2) the vertices are at V(0, a) and V 0 (0  a), the foci are at F(0, c) and F 0 (0, c), and the covertices are at B(b, 0) and B 0 (b, 0) (see Fig. 17-7). If the center of an ellipse is C(h, k) then the standard forms for the ellipses are:

(3)

(x  h)2 (y  k)2 þ ¼1 a2 b2

and

(4)

(y  k)2 (x  h)2 þ ¼1 a2 b2

In (3) the major axis is parallel to the x axis and the minor axis is parallel to the y axis. The foci have coordinates F(h þ c, k) and F 0 (h  c, k), the vertices are at V(h þ a, k) and V 0 (h  a, k), and the covertices are at B(h, k þ b) and B 0 (h, k  b) (see Fig. 17-8). In (4) the major axis is parallel to the y axis and the minor axis is parallel to the

Fig. 17-6

Fig. 17-7

CHAP. 17]

CONIC SECTIONS

175

x axis. The foci are at F(h, k þ c) and F 0 (h, k  c), the vertices have coordinates V(h, k þ a) and V 0 (h, k  a), and the covertices are at B(h þ b, k) and B0 (h  b, k) (see Fig. 17-9). EXAMPLES 17.8.

Determine the center, foci, vertices, and covertices for each ellipse.

(a)

x2 y2 þ ¼1 25 9

(c)

(x  3)2 (y  4)2 þ ¼1 225 289

(b)

x2 y2 þ ¼1 3 10

(d )

(x þ 1)2 (y  2)2 þ ¼1 100 64

(a)

x2 y2 þ ¼1 25 9

Fig. 17-8

Fig. 17-9

176

CONIC SECTIONS

[CHAP. 17

Since a2 is the greater denominator, a2 ¼ 25 and b2 ¼ 9, so a ¼ 5 and b ¼ 3. From a2 ¼ b2 þ c2 , we get 25 ¼ 9 þ c2 and c ¼ 4. The center is at (0, 0). The vertices are at (a, 0) and (a, 0), so V(5, 0) and V 0 (5, 0). The foci are at (c, 0) and (c, 0), so F(4, 0) and F(4, 0). The covertices are at (0, b) and (0, b), so B(0, 3) and B0 (0, 3). (b)

y2 x2 þ ¼1 10 3

pffiffiffiffiffi pffiffiffi pffiffiffi a2 ¼ 10 and b2 ¼ 3, so a ¼ 10, b ¼ 3, and since a2 ¼ b2 þ c2 , c ¼ 7. Since y2 is over the larger denominator, the vertices and foci are on the y axis. The center is (0, 0). pffiffiffiffiffi pffiffiffiffiffi vertices (0, a) and (0, a) V(0, 10), V 0 (0,  10) pffiffiffi pffiffiffi foci (0, c) and (0, c) F(0, 7), F 0 (0,  7) pffiffiffi p ffiffi ffi covertices (b, 0) and (b, 0) B( 3, 0), B0 ( 3, 0) (c)

(y  4)2 (x  3)2 þ ¼1 289 225 a2 ¼ 289 and b2 ¼ 225, so a ¼ 17 and b ¼ 15 and from a2 ¼ b2 þ c2 , c ¼ 8. Since ( y  4)2 is over a2 , the vertices and foci are on a line parallel to the y axis. center (h, k) ¼ (3, 4) vertices (h, k þ a) and (h, k 2 a) foci (h, k þ c) and (h, k 2 c) covertices (h þ b, k) and (h  b, k)

(d )

V(3, 21), V 0 (3, 13) F(3, 12), F 0 (3, 4) B(18, 4), B0 (12, 4)

(x þ 1)2 (y  2)2 þ ¼1 100 64 a2 ¼ 100, b2 ¼ 64, so a ¼ 10 and b ¼ 8. From a2 ¼ b2 þ c2 , we get c ¼ 6. Since (x þ 1)2 is over a2 the vertices and foci are on a line parallel to the x axis. center (h, k) ¼ (1, 2) vertices (h þ a, k) and (h  a, k) foci (h þ c, k) and (h c, k) covertices (h, k þ b) and (h, k  b)

V(9, 2), V0 (11, 2) F(5, 2), F0 (7, 2) B(1, 10), B0 (1, 6)

EXAMPLES 17.9. Write the equation of the ellipse having the given characteristics. (a) central ellipse, foci at (+4, 0), and vertices at (+5, 0) (b) center at (0, 3), major axis of length 12, foci at (0, 6) and (0, 0) (a)

A central ellipse has its center at the origin, so (h, k) ¼ (0, 0). Since the vertices are on the x axis and the center is at (0, 0), the form of the ellipse is x2 y2 þ ¼1 a2 b2 From a vertex at (5, 0) and the center at (0, 0), we get a ¼ 5. From a focus at (4, 0) and the center at (0, 0), we get c ¼ 4. Since a2 ¼ b2 þ c2 , 25 ¼ b2 þ 16, so b2 ¼ 9 and b ¼ 3. The equation of the ellipse is

(b)

x2 y2 þ ¼1 25 9

Since the center is at (0, 3), h ¼ 0 and k ¼ 3. Since the foci are on the y axis, the form of the equation of the ellipse is (y  k)2 (x  h)2 þ ¼1 a2 b2 The foci are (h, k þ c) and (h, k  c), so (0, 6) ¼ (h, k þ c) and 3 þ c ¼ 6 and c ¼ 3. The major axis length is 12, so we know 2a ¼ 12 and a ¼ 6.

CHAP. 17]

177

CONIC SECTIONS

From a2 ¼ b2 þ c2 , we get 36 ¼ b2 þ 9 and b2 ¼ 27. The equation of the ellipse is

(y  3)2 x2 þ ¼1 36 27

EXAMPLE 17.10. Write the equation of the ellipse 18x2 þ 12y2  144x þ 48y þ 120 ¼ 0 in standard form. 18x2 þ 12y2  144x þ 48y þ 120 ¼ 0 (18x2  144x) þ (12y2 þ 48y) ¼ 120 18(x2  8x) þ 12(y2 þ 4y) ¼ 120 18(x2  8x þ 16) þ 12(y2 þ 4y þ 4) ¼ 120 þ 18(16) þ 12(4)

reorganize terms factor to get x2 and y2 complete square on x and y

18(x  4)2 þ 12(y þ 2)2 ¼ 216

simplify

18(x  4) 12(y þ 2) þ ¼1 216 216 (x  4)2 (y þ 2)2 þ ¼1 12 18 2

17.6

2

divide by 216 standard form

HYPERBOLAS

The hyperbola is the locus of all points in a plane such that for any point of the locus the difference of the distances from two fixed points, the foci, is a constant. Central hyperbolas have their center at the origin and their vertices and foci on one axis, and are symmetric with respect to the other axis. The standard form equations for central hyperbolas are: (1)

x2 y2  ¼1 a2 b2

and

(2)

y2 x2  ¼1 a2 b2

The distance from the center to a vertex is denoted by a and the distance from the center to a focus is c. For a hyperbola, c2 ¼ a2 þ b2 and b is a positive number. The line segment between the vertices is called the transverse axis. The denominator of the positive fraction for the standard form is always a2 . 0 In (1) the transverse axis VV lies on the x axis, the vertices are V(a, 0) and V 0 ða, 0), and the foci are at 0 0 F(c, 0) and F (c, 0) (see Fig. 17-10). In (2) the transverse axis VV lies on the y axis, the vertices are at V(0, a) 0 0 and V (0, a), and the foci are at F(0, c) and F (0, c) (see Fig. 17-11). When lines are drawn through the points R and C and the points S and C, we have the asymptotes of the hyperbola. The asymptote is a line that the graph of the hyperbola approaches but does not reach. If the center of the hyperbola is at (h, k) the standard forms are (3) and (4): (3)

(x  h)2 (y  k)2  ¼1 a2 b2

and

(4)

(y  k)2 (x  h)2  ¼1 a2 b2

In (3) the transverse axis is parallel to the x axis, the vertices have coordinates V(h þ a, k) and V 0 (h  a, k), the foci have coordinates F(h þ c, k) and F 0 (h  c, k), and the points R and S have coordinates R(h þ a, k þ b) and S(h þ a, k  b). The lines through R and C and S and C are the asymptotes of the hyperbola (see Fig. 17-12). In equation (4) the transverse axis is parallel to the y axis, the vertices are at V(h, k þ a) and V 0 (h, k  a), the foci are at F(h, k þ c) and F 0 (h, k  c), and the points R and S have coordinates R(h þ b, k þ a) and S(h  b, k þ a) (see Fig. 17-13). EXAMPLES 17.11. Find the coordinates of the center, vertices, and foci for each hyperbola. (a)

(x  4)2 (y  5)2  ¼1 9 16

(b)

(y þ 5)2 (x þ 9)2  ¼1 25 144

(c)

(x þ 3)2 (y  4)2  ¼1 225 64

CHAP. 17]

CONIC SECTIONS

center C(h, k) ¼ (9, 5) vertices are V(h, k þ a) and V 0 (h, k  a) foci are F(h, k þ c) and F 0 (h, k  c) (c)

V(9, 0) and V 0 (9, 10) F(9, 8) and F 0 (9, 18)

(x þ 3)2 (y  4)2  ¼1 225 64 Since a2 = 225 and b2 = 64, we get a ¼ 15 and b ¼ 8. From c2 ¼ a2 þ b2 , we get c ¼ 17. center C(h, k) ¼ (3, 4) vertices are V(h þ a, k) and V 0 (h  a, k) foci are F(h þ c, k) and F 0 (h c, k)

V(12, 4) and V 0 (18, 4) F(14, 4) and F 0 (20, 4)

EXAMPLES 17.12. Write the equation of the hyperbola that has the given characteristics. (a) Foci are at (2, 5) and (4, 5) and transverse axis has length 4. (b) Center at (1, 3), a focus is at (1, 2) and a vertex is at (1, 1). (a) The foci are on a line parallel to the x axis, so the form is (x  h)2 (y  k)2  ¼1 a2 b2 The center is half-way between the foci, so c ¼ 3 and the center is at C(1, 5). The transverse axis joins the vertices, so its length is 2a, so 2a ¼ 4 and a ¼ 2. Since c2 ¼ a2 þ b2 , c ¼ 3 and a ¼ 2, so b2 ¼ 5. The equation of the hyperbola is (x þ 1)2 (y  5)2  ¼1 4 5 (b) The distance from the vertex (1, 1) to the center (1, 3) is a, so a ¼ 4. The distance from the focus (1, 2) to the center (1, 3) is c, so c ¼ 5. Since c2 ¼ a2 þ b2 , a ¼ 4, and c ¼ 5, b2 ¼ 9. Since the center, vertex, and focus lie on a line parallel to the y axis, the hyperbola has the form (y  k)2 (x  h)2  ¼1 a2 b2 The center is (1, 3), so h ¼ 1, and k ¼ 3. The equation of the hyperbola is (y þ 3)2 (x  1)2  ¼1 16 9 EXAMPLES 17.13. Write the equation of each hyperbola in standard form. (a) 25x2  9y2  100x  72y  269 ¼ 0

(b) 4x2  9y2  24x  90y  153 ¼ 0

(a) 25x2  9y2  100x  72y  269 ¼ 0 (25x2  100x) þ (9y2  72y) ¼ 269 25(x2  4x)  9(y2 þ 8y) ¼ 269 25(x2  4x þ 4)  9(y2 þ 8y þ 16) ¼ 269 þ 25(4)  9(16) 25(x  2)2  9(y þ 4)2 ¼ 225 (x  2)2 (y þ 4)2  ¼1 9 25

rearrange terms factor to get x2 and y2 complete square for x and y simplify then divide by 225 standard form

179

180

CONIC SECTIONS

[CHAP. 17

(b) 4x2  9y2  24x  90y  153 ¼ 0 (4x2  24x) þ (9y2  90y) ¼ 153 4(x2  6x)  9(y2 þ 10y) ¼ 153 4(x2  6x þ 9)  9(y2 þ 10y þ 25) ¼ 153 þ 4(9)  9(25) 4(x  3)2  9(y þ 5)2 ¼ 36

reorganize terms factor to get x2 and y2 complete square for x and y simplify then divide by 36

(x  3)2 (y þ 5)2  ¼1 9 4 2 2 (y þ 5) (x  3)  ¼1 4 9

17.7

simplify signs standard form

GRAPHING CONIC SECTIONS WITH A CALCULATOR

Since most conic sections are not functions, an important step is to solve the standard form equation for y. If y is equal to an expression in x that contains a + quantity, we need to separate the expression into two parts: y1 ¼ the expression using the þ quantity and y2 ¼ the expression using the  expression. Otherwise, set y1 ¼ the expression. Graph either y1 or y1 and y2 simultaneously. The window may need to be adjusted to correct for the distortion caused by unequal scales used on the x axis and the y axis in many graphing calculators’ standard windows. Setting the y scale to 0.67 often corrects for this distortion. For the circle, ellipse, and hyperbola, it is usually necessary to center the graphing window at the point (h, k), the center of the conic section. However, the parabola is viewed better if the vertex (h, k) is at one end of the viewing window.

Solved Problems 17.1

Draw the graph of each of the following equations: (a) 4x2 þ 9y2 ¼ 36,

(b) 4x2  9y2 ¼ 36,

(c) 4x þ 9y2 ¼ 36:

SOLUTION (a)

4 2pffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ + 9  x2 . (9  x2 ), 9 3 Note that y is real when 9  x2  0, i.e., when 3  x  3. Hence values of x greater than 3 or less than 3 are excluded. 4x2 þ 9y2 ¼ 36,

y2 ¼

x

3

2

1

0

1

2

3

y

0

+1:49

+1:89

+2

+1:89

+1:49

0

The graph is an ellipse with center at the origin (see Fig. 17-14(a)).

Fig. 17-14

CHAP. 17]

181

CONIC SECTIONS

(b) 4x2  9y2 ¼ 36,

y2 ¼

ffi 2 pffiffiffiffiffiffiffiffiffiffiffiffi y¼+ x2  9. 3

4 2 (x  9), 9

Note that x cannot have a value between 3 and 3 if y is to be real. x

6

5

4

+3:46 +2:67 +1:76

y

3

3

0

0

4

5

6

+1:76 +2:67 +3:46

The graph consists of two branches and is called a hyperbola (see Fig. 17-14(b)). 4 2pffiffiffiffiffiffiffiffiffiffiffi y ¼ + 9  x: y2 ¼ (9  x), (c) 4x þ 9y2 ¼ 36, 9 3 Note that if x is greater than 9, y is imaginary. x

1

0

1

5

8

9

y

+2:11

+2

+1:89

+1:33

+0:67

0

The graph is a parabola (see Fig. 17-14(c)).

17.2

Plot the graph of each of the following equations: (a) xy ¼ 8,

(b) 2x2  3xy þ y2 þ x  2y  3 ¼ 0,

(c) x2 þ y2  4x þ 8y þ 25 ¼ 0.

SOLUTION (a) xy ¼ 8, y ¼ 8/x. Note that if x is any real number except zero, y is real. The graph is a hyperbola (see Fig. 17-15(a)). x

4

2

1

1 2

 12

1

2

4

y

2

4

8

16

16

8

4

2

Fig. 17-15 (b) 2x2  3xy þ y2 þ x  2y  3 ¼ 0. Write as y2  (3x þ 2)y þ (2x2 þ x  3) ¼ 0 and solve by the quadratic formula to obtain

y¼

3x þ 2 +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ 8x þ 16 (3x þ 2) + (x þ 4) ¼ 2 2

or

y ¼ 2x þ 3, y ¼ x  1

182

CONIC SECTIONS

[CHAP. 17

The given equation is equivalent to two linear equations, as can be seen by writing the given equation as (2x  y þ 3)(x  y  1) ¼ 0. The graph consists of two intersecting lines (see Fig. 17-15(b)). (c) Write as y2 þ 8y þ (x2  4x þ 25) ¼ 0; solving, y¼

4 +

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4(x2  4x þ 9) : 2

Since x2  4x þ 9 ¼ x2  4x þ 4 þ 5 ¼ (x  2)2 þ 5 is always positive, the quantity under the radical sign is negative. Thus y is imaginary for all real values of x and the graph does not exist.

17.3

For each equation of a circle, write it in standard form and determine the center and radius. (b) 4x2 þ 4y2 þ 28y þ 13 ¼ 0 (a) x2 þ y2  8x þ 10y  4 ¼ 0 SOLUTION (a) x2 þ y2  8x þ 10y  4 ¼ 0 (x2  8x þ 16) þ (y2 þ 10y þ 25) ¼ 4 þ 16 þ 25 (x  4)2 þ (y þ 5)2 ¼ 45 center: C(4, 5)

standard form pffiffiffiffiffi pffiffiffi radius: r ¼ 45 ¼ 3 5

(b) 4x2 þ 4y2 þ 28y þ 13 ¼ 0 x2 þ y2 þ 7y ¼ 13=4 x2 þ (y2 þ 7y þ 49=4) ¼ 13=4 þ 49=4 x2 þ (y þ 7=2)2 ¼ 9 center: C(0, 7=2)

17.4

standard form radius: r ¼ 3

Write the equation of the following circles. (a) center at the origin and goes through (2, 6) (b) ends of diameter at (7, 2) and (5, 4) SOLUTION (a) The standard form of a circle with center at the origin is x2 þ y2 ¼ r 2 . Since the circle goes through (2, 6), we substitute x ¼ 2 and y ¼ 6 to determine r 2 . Thus, r 2 ¼ 22 þ 62 ¼ 40. The standard form of the circle is x2 þ y2 ¼ 40. (b) The center of a circle is the midpoint of the diameter. The midpoint M of the line segment having endpoints (x1 , y1 ) and (x2 , y2 ) is M¼

! x1 þ x2 y1 þ y2 , : 2 2

Thus, the center is C

    7 þ 5 2 þ 4 2 6 , ¼C , ¼ C(1, 3): 2 2 2 2

The radius of a circle is the distance from the center to the endpoint of the diameter. The distance, d, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ) and (x2 , y2 ) is d ¼ (x2  x1 )2 þ (y2  y1 )2 . Thus, the distance from the center between two points (x1 ,p pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi C(l, 3) to (5, 4) is r ¼ (5  (1))2 þ (4  3)2 ¼ 62 þ 12 ¼ 37. The equation of the circle is (x þ 1)2 þ (y  3)2 ¼ 37.

17.5

Write the equation of the circle passing through three points (3, 2), (1, 4), and (2, 3).

CHAP. 17]

183

CONIC SECTIONS

SOLUTION The general form of the equation of a circle is x2 þ y2 þ Dx þ Ey þ F ¼ 0, so we must substitute the given points into this equation to get a system of equations in D, E, and F. For (3, 2) For (1, 4) For (2, 3)

32 þ 22 þ D(3) þ E(2) þ F ¼ 0 (21) þ 42 þ D(1) þ E(4) þ F ¼ 0 22 þ 32 þ D(2) þ E(3) þ F ¼ 0 2

3D þ 2E þ F ¼ 13 D þ 4E þ F ¼ 17 2D þ 3E þ F ¼ 13

then (1) then (2) then (3)

We eliminate F from (1) and (2) and from (1) and (3) to get (4) 4D  2E ¼ 4

and

(5) D  E ¼ 0:

We solve the system of (4) and (5) to get D ¼ 2 and E ¼ 2 and substituting into (1) we get F ¼ 23. The equation of the circle is x2 þ y2 þ 2x þ 2y  23 ¼ 0.

17.6

Write the equation of the parabola in standard form and determine the vertex, focus, directrix, and axis. (a) y2  4x þ 10y þ 13 ¼ 0

(b) 3x2 þ 18x þ 11y þ 5 ¼ 0:

SOLUTION (a) y2  4x þ 10y þ 13 ¼ 0 y2 þ 10y ¼ 4x  13 y2 þ 10y þ 25 ¼ 4x þ 12 ( y þ 5)2 ¼ 4(x þ 3) vertex (h, k) ¼ (3, 5) focus (h þ p, k) ¼ (3 þ 1, 5) ¼ (2, 5) directrix: x ¼ h  p ¼ 4

rearrange terms complete the square for y standard form 4p ¼ 4 so p ¼ 1 axis: y ¼ k ¼ 5

(b) 3x2 þ 18x þ 11y þ 5 ¼ 0 x2 þ 6x ¼ 11=3y  5=3 x2 þ 6x þ 9 ¼ 11=3y þ 22=3 (x þ 3)2 ¼ 11=3(y  2) vertex (h, k) ¼ (3; þ2) focus (h, k þ p) ¼ (3, þ2 þ (11=12)) ¼ (3, 13=12) directrix ¼ k  p ¼ þ2  (11=12) ¼ 35=12

17.7

standard form 4p ¼ 11=3 p ¼ 11=12 axis: x ¼ h ¼ 3

Write the equation of the parabola with the given characteristics. (a) vertex at origin and directrix y ¼ 2 (b) vertex (1, 3) and focus (3, 3) SOLUTION (a) Since the vertex is at the origin, we have the form y2 ¼ 4px or x2 ¼ 4py. However, since the directrix is y ¼ 2, the form is x2 ¼ 4py. The vertex is (0, 0) and the directrix is y ¼ k  p. Since y ¼ 2 and k ¼ 0, we have p ¼ 2. The equation of the parabola is x2 ¼ 8y: (b) The vertex is (1, 3) and the focus is (3, 3) and since they lie on a line parallel to the x axis, the standard form is (y  k)2 ¼ 4p(x  h): From the vertex we get h ¼ 1 and k ¼ 3, and since the focus is (h þ p, k), h þ p ¼ 3 and 1 þ p ¼ 3, we get p ¼ 2. Thus, the standard form of the parabola is (y þ 3)2 ¼ 8(x þ 1).

17.8

Write the equation of the ellipse in standard form and determine its center, vertices, foci, and covertices. (a) 64x2 þ 81y2 ¼ 64

(b) 9x2 þ 5y2 þ 36x þ 10y  4 ¼ 0

184

CONIC SECTIONS

[CHAP. 17

SOLUTION (a)

64x2 þ 81y2 ¼ 64 81y2 ¼1 x2 þ 64 x2 y2 þ ¼1 64 1 81 center is the origin (0, 0)

divide by 64 divide the numerator and denominator by 81 standard form a2 ¼ 1 and b2 ¼ 64=81, so a ¼ 1 and b ¼ 8=9

For an ellipse, a2 ¼ b2 þ c2 , so 1 ¼ 64=81 þ c2 and c2 ¼ 17=81, giving c ¼ 0 The vertices are (a, 0) and (a, 0), sopV(1, 0).ffiffiffiffiffi ffiffiffiffiffi 0) and V (1, p The foci are (c, 0), and (c, 0), so F( 17=9, 0) and F0 ( 17=9, 0). The covertices are (0, b) and (0, b), so B(0, 8/9) and B0 (0, 8=9). (b)

17.9

pffiffiffiffiffi 17=9.

9x2 þ 5y2 þ 36x þ 10y  4 ¼ 0 9(x2 þ 4x þ 4) þ 5(y2 þ 2y þ 1) ¼ 4 þ 36 þ 5 9(x þ 2)2 þ 5(y þ 1)2 ¼ 45 (x þ 2)2 (y þ 1)2 þ ¼1 standard form 5 9 pffiffiffi center (h, k) ¼ (2, 1) a2 ¼ 9, b2 ¼ 5, so a ¼ 3 and b ¼ 5 Since a2 ¼ b2 þ c2 , c2 ¼ 4 and c ¼ 2: The vertices are (h, k þ a) and (h, k  a), so V(22, 2) and V 0 (22, 24). The foci are (h, k þ c) and (h, k  c), so F(2, 1) and Fp0 (2,  3). ffiffiffi pffiffiffi The covertices are (h þ b, k) and (h  b, k) so B(2 þ 5, 1) and B 0 (2  5, 1).

Write the equation of the ellipse that has these characteristics. pffiffiffi (a) foci are (1, 0) and (1, 0) and length of minor axis is 2 2. (b) vertices are at (5, 1) and (3, 1) and c ¼ 3. SOLUTION (a)

The midpoint of the line segment between the foci is the center, so the center is C(0, 0) and we have a central ellipse. The standard form is x2 y2 þ ¼1 a2 b2

or

y2 x2 þ ¼1 a2 b2

The foci are (c, 0) and (c, p 0)ffiffiffiso (c, 0) ¼ p (1,ffiffiffi0) and c ¼p1. ffiffiffi The minor axis has length 2 2, so 2b ¼ 2 2 and b ¼ 2 and b2 ¼ 2: 2 2 2 2 For the ellipse, a ¼ b þ c and a ¼ 1 þ 2 ¼ 3. Since the foci are on the x axis, the standard form is x2 y2 þ ¼1 a2 b2 The equation of the ellipse is x2 y2 þ ¼ 1: 3 2 (b)

The midpoint of the line segment between the vertices is the center, so the center is   5  3 1  1 C , ¼ (1; 21): 2 2 We have an ellipse with center at (h, k) where h ¼ 1 and k ¼ 1.

CHAP. 17]

CONIC SECTIONS

185

The standard form of the ellipse is (x  h)2 (y  k)2 þ ¼1 a2 b2

or

(y  k)2 (x  h)2 þ ¼ 1: a2 b2

The vertices are (h þ a, k) and (h  a, k), so (h þ a, k) ¼ (1 þ a, 1) ¼ (5, 1). Thus, 1 þ a ¼ 5 and a ¼ 4. For the ellipse, a2 ¼ b2 þ c2 , c is given to be 3, and we found a to be 4. Thus, a2 ¼ 42 ¼ 16 and 2 c ¼ 32 ¼ 9. Therefore, a2 ¼ b2 þ c2 yields 16 ¼ b2 þ 9 and b2 ¼ 7. Since the vertices are on a line parallel to the x axis, the standard form is (x  h)2 (y  k)2 þ ¼ 1: a2 b2 The equation of the ellipse is (x  1)2 (y þ 1)2 þ ¼ 1: 16 7

17.10 For each hyperbola, write the equation in standard form and determine the center, vertices, and foci. (a)

16x2  9y2 þ 144 ¼ 0

(b)

9x2  16y2 þ 90x þ 64y þ 17 ¼ 0

SOLUTION (a)

16x2  9y2 þ 144 ¼ 0 16x2  9y2 ¼ 144 x2 y2  ¼1 9 16 2 2 y x  ¼1 16 9

standard form

center (h, k) ¼ (0, 0) a2 ¼ 16 and b2 ¼ 9, so a ¼ 4 and b ¼ 3 2 2 2 Since c ¼ a þ b for a hyperbola, c2 ¼ 16 þ 9 ¼ 25 and c ¼ 5. The foci are (0, c) and (0, c), so F(0, 5) and F 0 (0, 5). The vertices are (0, a) and (0, a), so V(0, 4) and V 0 (0, 4) (b)

9x2  16y2 þ 90x þ 64y þ 17 ¼ 0 9(x2 þ 10x þ 25)  16(y2  4y þ 4) ¼ 17 þ 225  64 9(x þ 5)2  16(y  2)2 ¼ 144 (x þ 5)2 (y  2)2  ¼1 16 9 center (h, k) ¼ (25, 2)

standard form a 2 ¼ 16 and b 2 ¼ 9, so a ¼ 4 and b ¼ 3

Since c2 ¼ a2 þ b2 , c2 ¼ 16 þ 9 ¼ 25 and c ¼ 5: The foci are (h þ c, k) and (h 2 c, k), so F(0, 2) and F 0 (10, 2): The vertices are (h þ a, k) and (h  a, k), so V(21, 2) and V 0 (29, 2).

17.11 Write the equation of the hyperbola with the given characteristics. (a) (b)

vertices are (0, +2) and foci are (0, +3) foci (1, 2) and (11, 2) and the transverse axis has length 4

186

CONIC SECTIONS

[CHAP. 17

SOLUTION (a)

Since the vertices are (0, +2), the center is at (0, 0), and since they are on a vertical line the standard form is y2 x2  ¼1 a2 b2 The vertices are at (0, +a) so a ¼ 2 and the foci are at (0, +3) so c ¼ 3. Since c2 ¼ a2 þ b2 , 9 ¼ 4 þ b2 so b2 ¼ 5. The equation of the hyperbola is y2 x2  ¼1 4 5

(b)

Since the foci are (1, 2) and (11, 2), they are on a line parallel to the x axis, so the form is (x  h)2 (y  k)2  ¼1 a2 b2 The midpoint of the line segment between the foci (1, 2) and (11, 2) is the center, so C(h, k) = (5, 2). The foci are at (h þ c, k) and (h  c, k), so (h þ c, k) ¼ (1, 2) and 5 þ c ¼ 1, with c ¼ 6. The transverse axis has length 4 so 2a ¼ 4 and a ¼ 2. From c2 ¼ a2 þ b2 , we get 36 ¼ 4 þ b2 and b2 ¼ 32. The equation of the hyperbola is (x þ 5)2 (y  2)2  ¼1 4 32

Supplementary Problems 17.12

Graph each of the following equations. (a) (b) (c) (d )

17.13

x2 þ y2 ¼ 9 xy ¼ 4 4x2 þ y2 ¼ 16 x2  4y2 ¼ 36

(e) (f) (g) (h)

y2 ¼ 4x x2 þ 3y2  1 ¼ 0 x2 þ 3xy þ y2 ¼ 16 x2 þ 4y ¼ 4

Write the equation of the circle that has the given characteristics. (a) center (4, 1) and radius 3 (b) center (5, 3) and radius 6

17.14

(c) goes through (0, 0), (4, 0), and (0, 6) (d ) goes through (2, 3), (1, 7), and (1, 5)

Write the equation of the circle in standard form and state the center and radius. (a) x2 þ y2 þ 6x  12y  20 ¼ 0 (b) x2 þ y2 þ 12x  4y  5 ¼ 0

17.15

(c) x2 þ y2 þ 7x þ 3y  10 ¼ 0 (d ) 2x2 þ 2y2  5x  9y þ 11 ¼ 0

Write the equation of the parabola that has the given characteristics. (a) (b) (c) (d )

17.16

(i) x2 þ y2  2x þ 2y þ 2 ¼ 0 ( j) 2x2  xy  y2  7x  2y þ 3 ¼ 0

vertex (3, 22) and directrix x ¼ 5 vertex (3, 5) and focus (3, 10) passes through (5, 10), vertex is at the origin, and axis is the x axis vertex (5, 4) and focus (2, 4)

Write the equation of the parabola in standard form and determine its vertex, focus, directrix, and axis. (a) y2 þ 4x  8y þ 28 ¼ 0 (b) x2  4x þ 8y þ 36 ¼ 0

(c) y2  24x þ 6y  15 ¼ 0 (d ) 5x2 þ 20x  9y þ 47 ¼ 0

CHAP. 17]

187

CONIC SECTIONS

17.17

Write the equation of the ellipse that has these characteristics. pffiffiffi (a) vertices (+4, 0), foci (+2 3, 0) (b) covertices (+3, 0), major axis length 10 (c) center (3, 2), vertex (2, 2), c ¼ 4 (d ) vertices (3, 2) and (3, 6), covertices (1, 2) and (5, 2)

17.18

Write the equation of the ellipse in standard form and determine the center, vertices, foci, and covertices. (a) 3x2 þ 4y2  30x  8y þ 67 ¼ 0 (b) 16x2 þ 7y2  64x þ 28y  20 ¼ 0

17.19

Write the equations of the hyperbola that has the given characteristics. (a) (b) (c) (d )

17.20

(c) 9x2 þ 8y2 þ 54x þ 80y þ 209 ¼ 0 (d ) 4x2 þ 5y2  24x  10y þ 21 ¼ 0

vertices (+3, 0), foci (+5, 0) vertices (0, +8), foci (0, +10) foci (4, 1) and (4, 5), transverse axis length is 2 vertices (1, 1) and (1, 5), b ¼ 5

Write the equation of the hyperbola in standard form and determine the center, vertices, and foci. (a) 4x2  5y2  8x  30y  21 ¼ 0 (b) 5x2  4y2  10x  24y  51 ¼ 0

(c) 3x2  y2  18x þ 10y  10 ¼ 0 (d ) 4x2  y2 þ 8x þ 6y þ 11 ¼ 0

ANSWERS TO SUPPLEMENTARY PROBLEMS 17.12

(a) (b) (c) (d ) (e)

circle, Fig. 17-16 hyperbola, Fig. 17-17 ellipse, Fig. 17-18 hyperbola, Fig. 17-19 parabola, Fig. 17-20

17.13

(a) (x  4)2 þ (y  1)2 ¼ 9 (b) (x  5)2 þ (y þ 3)2 ¼ 36

Fig. 17-16

(f) (g) (h) (i) ( j)

ellipse, Fig. 17-21 hyperbola, Fig. 17-22 parabola, Fig. 17-23 single point, (1, 1) two intersecting lines, Fig. 17-24 (y ¼ x  3 and y ¼ 2x þ 1)

(c) x2 þ y2 þ 4x  6y ¼ 0 (d ) x2 þ y2 þ 11y  y  32 ¼ 0

Fig. 17-17

188

CONIC SECTIONS

Fig. 17-18

[CHAP. 17

Fig. 17-19

Fig. 17-20 Fig. 17-21

Fig. 17-22

Fig. 17-23

CHAP. 17]

189

CONIC SECTIONS

Fig. 17-24

pffiffiffiffiffi (x þ 3)2 þ (y  6)2 ¼ 65, C(23, 6), r ¼ 65 pffiffiffi (x þ 6)2 þ (y  2)2 ¼ 45, C( 6, 2), r ¼ 3 5 pffiffiffi C(7=2, 3=2), r ¼ 7 2=2 (x þ 7=2)2 þ (y þ 3=2)2 ¼ 49=2, p ffiffi ffi (x  5=4)2 þ (y  9=4)2 ¼ 9=8, C(5/4, 9/4), r ¼ 3 2=4

17.14

(a) (b) (c) (d )

17.15

(a) (y þ 2)2 ¼ 32(x  3)

17.16

(a) (b) (c) (d )

(y  4)2 ¼ 4(x þ 3), (x  2)2 ¼ 8(y þ 4), (y þ 3)2 ¼ 24(x þ 1), (x þ 2)2 ¼ 9(y  3)=5,

17.17

(a)

x2 y2 þ ¼1 16 4

(c)

(x þ 3)2 (y  2)2 þ ¼1 25 9

(b)

y2 x2 þ ¼1 25 9

(d )

(y þ 2)2 (x  3)2 þ ¼1 16 4

(b) (x  3)2 ¼ 20(y  5) V(3, 4), V(2, 4), V(1, 3), V(2, 3),

(c) y2 ¼ 20x

(d) (x  5)2 ¼ 12(y  4)

F(4, 4), directrix: x ¼ 2, axis: y ¼ 4 F(2, 6), directrix: y ¼ 2, axis: x ¼ 2 F(5, 3), directrix: x ¼ 7, axis: y ¼ 3 F(2, 69=20), directrix: y ¼ 51=20, axis: x ¼ 2

(x  5)2 (y  1)2 þ ¼ 1, center (5, 1), vertices 1);ffiffiffi foci (6, 1) and (4, 1), pffiffiffi (7, 1) and (3,p 4 3 covertices (5, 1 þ 3) and (5, 1  3) (y þ 2)2 (x  2)2 (b) þ ¼ 1, center (2, 2), vertices (2, 2) and p (2,ffiffiffi 6); foci (2, 1) and (2, 5), pffiffiffi 16 7 covertices (2 þ 7, 2) and (2  7, 2) (y þ 5)2 (x þ 3)2 þ ¼ 1, center (3, 5), vertices (3,p2) (c) ffiffiffi and (3, 8), foci pffiffiffi (3, 4) and 9 8 (3, 6), covertices (3 þ 2 2, 5) and (32 2; 5) pffiffiffi pffiffiffi (x  3)2 (y  1)2 (d ) þ ¼ 1, center (3, 1), vertices (3 þ 5, 1) and (3  5, 1), foci (4, 1) and (2, 1), 5 4 covertices (3, 3) and (3, 1)

17.18

(a)

17.19

(a)

x2 y2  ¼1 9 16

(c)

(y  2)2 (x  4)2  ¼1 1 8

(b)

y2 x2  ¼1 64 36

(d )

(y  2)2 (x þ 1)2  ¼1 9 25

190

17.20

CONIC SECTIONS

[CHAP. 17

(y þ 3)2 (x  1)2  ¼ 1, center (1, 3), vertices (1, 1) and (1, 5), foci (1, 0) and (1, 8) 4 5 (x  1)2 (y  3)2  ¼ 1, center (1, 3), vertices (1, 3) and (3, 3), foci (4, 3) and (2, 3) (b) 4 5 2 2 (x  3) (y þ 5)  ¼ 1, center (3,5), vertices (5, 5) and (1, 5), foci (7, 5) and (1, 5) (c) 4 12 2 2 pffiffiffi pffiffiffi (y  3) (x þ 1)  ¼ 1, center (1, 3), vertices (1, 7) and (1, 1), foci (1, 3 þ 2 5) and (1, 3  2 5) (d) 16 4 (a)

CHAPTER 18

Systems of Equations Involving Quadratics 18.1

GRAPHICAL SOLUTION

The real simultaneous solutions of two quadratic equations in x and y are the values of x and y corresponding to the points of intersection of the graphs of the two equations. If the graphs do not intersect, the simultaneous solutions are imaginary. 18.2 A.

ALGEBRAIC SOLUTION One linear and one quadratic equation Solve the linear equation for one of the unknowns and substitute in the quadratic equation. EXAMPLE 18.1.

Solve the system

(1) x þ y ¼ 7 (2) x2 þ y2 ¼ 25 Solving (1) for y, y ¼ 7 2 x. Substitute in (2) and obtain x 2 þ (7 2 x)2 ¼ 25, x 2 2 7x þ 12 ¼ 0, (x 2 3)(x 2 4) ¼ 0, and x ¼ 3, 4. When x ¼ 3, y ¼ 7 2 x ¼ 4; when x ¼ 4, y ¼ 7 2 x ¼ 3. Thus the simultaneous solutions are (3, 4) and (4, 3).

B.

Two equations of the form ax 2 þ by 2 ¼ c Use the method of addition or subtraction. EXAMPLE 18.2.

Solve the system

(1) 2x2  y2 ¼ 7 (2) 3x2 þ 2y2 ¼ 14 To eliminate y, multiply (1) by 2 and add to (2); then 7x2 ¼ 28,

x2 ¼ 4

Now put x ¼ 2 or x ¼ 22 in (1) and obtain y ¼ +1.

191

and

x ¼ +2:

192

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

[CHAP. 18

The four solutions are: (2, 1);

C.

(2, 1);

(2, 1);

(2, 1)

Two equations of the form ax 2 þ bxy þ cy 2 ¼ d EXAMPLE 18.3.

Solve the system

(1) x2 þ xy ¼ 6 (2) x2 þ 5xy  4y2 ¼ 10 Method 1. Eliminate the constant term between both equations. Multiply (1) by 5, (2) by 3, and subtract; then x2  5xy þ 6y2 ¼ 0, (x  2y)(x  3y) ¼ 0, x ¼ 2y and x ¼ 3y: Now put x ¼ 2y in (1) or (2) and obtain y 2 ¼ 1, y ¼ +1. When y ¼ 1, x ¼ 2y ¼ 2; when y ¼ 21, x ¼ 2y ¼ 22. Thus two solutions are: x ¼ 2, y ¼ 1; x ¼ 22, y ¼ 21. Then put x ¼ 3y in (1) or (2) and get 1 y2 ¼ , 2

y¼+

pffiffiffi 2 : 2

When y¼

pffiffiffi 2 , 2

x ¼ 3y ¼

pffiffiffi 3 2 ; 2

when y¼

pffiffiffi 2 , 2

x¼

pffiffiffi 3 2 : 2

Thus the four solutions are: (2, 1);

(2, 1);

 pffiffiffi pffiffiffi 2 3 2 ; , 2 2



pffiffiffi pffiffiffi 2 3 2  ; 2 2

Method 2. Let y ¼ mx in both equations. From (1): x2 þ mx2 ¼ 6,

x2 ¼

6 . 1þm

From (2): x2 þ 5mx2  4m2 x2 ¼ 10,

x2 ¼

10 : 1 þ 5m  4m2

Then 6 10 ¼ 1 þ m 1 þ 5m  4m2 from which m ¼ 12 , 13 ; hence y ¼ x/2, y ¼ x/3. The solution proceeds as in Method 1.

D.

Miscellaneous methods (1)

Some systems of equations may be solved by replacing them by equivalent and simpler systems (see Problems 18.8– 18.10).

CHAP. 18]

(2)

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

An equation is called symmetric in x and y if interchange of x and y does not change the equation. Thus x 2 þ y 2 2 3xy þ 4x þ 4y ¼ 8 is symmetric in x and y. Systems of symmetric equations may often be solved by the substitutions x ¼ u þ v, y ¼ u 2 v (see Problems 18.11– 18.12).

Solved Problems 18.1

Solve graphically the following systems: (a)

x2 þ y2 ¼ 25 , x þ 2y ¼ 10

(b)

x2 þ 4y2 ¼ 16 , xy ¼ 4

(c)

x2 þ 2y ¼ 9 2x2  3y2 ¼ 1

SOLUTION See Fig. 18-1.

Fig. 18-1

18.2

Solve the following systems: (a)

x þ 2y ¼ 4 y2  xy ¼ 7

(b)

3x  1 þ 2y ¼ 0 3x2  y2 þ 4 ¼ 0

SOLUTION (a) Solving the linear equation for x, x ¼ 4 2 2y. Substituting in the quadratic equation, y2  y(4  2y) ¼ 7, 3y2  4y  7 ¼ 0, (y þ 1)(3y  7) ¼ 0 and y ¼ 1, 7=3: If y ¼ 21, x ¼ 4 2 2y ¼ 6; if y ¼ 7/3, x ¼ 4 2 2y ¼ 22/3. The solutions are (6, 21) and (22/3, 7/3). (b) Solving the linear equations for y, y ¼ 12 (1  3x). Substituting in the quadratic equation, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 + 22  4(1)(5) 3x2  ½12(1  3x)2 þ 4 ¼ 0, x2 þ 2x þ 5 ¼ 0 and x ¼ ¼ 1 + 2i: 2(1) If x ¼ 1 þ 2i, y ¼ 12 (1 3x) ¼ 12 ½1  3(1 þ 2i) ¼ 12 (4  6i) ¼ 2  3i: If x ¼ 1 2i, y ¼ 12 (1 3x) ¼ 12 ½1  3(1 2i) ¼ 12 (4 þ 6i) ¼ 2 þ 3i: The solutions are (1 þ 2i, 2  3i) and (1 2i, 2 þ 3i):

18.3

193

Solve the system: (1) 2x2  3y2 ¼ 6,

(2) 3x2 þ 2y2 ¼ 35:

194

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

[CHAP. 18

SOLUTION To eliminate y, multiply (1) by 2, (2) by 3 and add; then 13x 2 ¼ 117, x 2 ¼ 9, x ¼ +3. Now put x ¼ 3 or x ¼ 23 in (1) and obtain y ¼ +2. The solutions are: (3, 2); (23, 2); (3, 22); (23, 22).

18.4

Solve the system: 8 3 (1) 2  2 ¼ 5, x y

5 2 þ 2 ¼ 38: 2 x y

(2)

SOLUTION The equations are quadratic in

1 1 1 1 and . Substituting u ¼ and v ¼ , we obtain x y x y 8u2  3v2 ¼ 5

5u2 þ 2v2 ¼ 38:

and

Solving simultaneously, u 2 ¼ 4, v2 ¼ 9 or x 2 ¼ 1/4, y 2 ¼ 1/9; then x ¼ +1/2, y ¼ +1/3. The solutions are: 

18.5

 1 1 , ; 2 3

  1 1  , , 2 3

Solve the system



 1 1 , ; 2 3



 1 1  , : 2 3

(1) 5x2 þ 4y2 ¼ 48 (2) x2 þ 2xy ¼ 16

by eliminating the constant terms. SOLUTION Multiply (2) by 3 and subtract from (1) to obtain 2x2  6xy þ 4y2 ¼ 0,

x2  3xy þ 2y2 ¼ 0,

(x  y)(x  2y) ¼ 0 and

x ¼ y,

x ¼ 2y:

16 4 pffiffiffi and y ¼ + 3. 3 3pffiffiffi Substituting x ¼ 2y in (1) or (2), we have y 2 ¼ 2 and y ¼ + 2. The four solutions are: Substituting x ¼ y in (1) or (2), we have y2 ¼

 pffiffiffi pffiffiffi 4 3 4 3 , ; 3 3

18.6

pffiffiffi  pffiffiffi 4 3 4 3  ,  ; 3 3

Solve the system by using the substitution y ¼ mx.

pffiffiffi pffiffiffi (2 2, 2);

pffiffiffi pffiffiffi (2 2,  2):

(1) 3x2  4xy ¼ 4 (2) x2  2y2 ¼ 2

SOLUTION Put y ¼ mx in (1); then 3x2  4mx2 ¼ 4

and

Put y ¼ mx in (2); then x2  2m2 x2 ¼ 2

and

Thus

4 2 , ¼ 3  4m 1  2m2

4m2  4m þ 1 ¼ 0,

4 . 3  4m 2 x2 ¼ . 1  2m2 x2 ¼

(2m  1)2 ¼ 0

and

Now substitute y ¼ mx ¼ 12x in (1) or (2) and obtain x2 ¼ 4, x ¼ +2. The solutions are (2, 1) and (22, 21).

1 1 m¼ , . 2 2

CHAP. 18]

18.7

195

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

Solve the system: (1) x2 þ y2 ¼ 40,

(2) xy ¼ 12:

SOLUTION From (2), y ¼ 12/x; substituting in (1), we have x2 þ

144 ¼ 40, x2

x4  40x2 þ 144 ¼ 0,

(x2  36)(x2  4) ¼ 0

x ¼ +6,

and

+2:

For x ¼ +6, y ¼ 12/x ¼ +2; for x ¼ +2, y ¼ +6. The four solutions are: (6, 2); (26, 22); (2, 6); (22, 26). Note. Equation (2) indicates that those solutions in which the product xy is negative (e.g. x ¼ 2, y ¼ 26) are extraneous.

18.8

Solve the system: (1) x2 þ y2 þ 2x  y ¼ 14,

(2) x2 þ y2 þ x  2y ¼ 9.

SOLUTION Subtract (2) from (1): x þ y ¼ 5 or y ¼ 5 2 x. Substitute y ¼ 5 2 x in (1) or (2): 2x 2 2 7x þ 6 ¼ 0, (2x 2 3)(x 2 2) ¼ 0 and x ¼ 3/2, 2. The solutions are ( 32 , 72 ) and (2, 3).

18.9

Solve the system: (1) x3 þ y3 ¼ 35,

(2) x þ y ¼ 5:

SOLUTION Dividing (1) by (2), x3 þ y3 35 ¼ xþy 5

and

(3) x2  xy þ y2 ¼ 7:

From (2), y ¼ 5 2 x; substituting in (3), we have x2  5x þ 6 ¼ 0, x2  x(5  x) þ (5  x)2 ¼ 7, The solutions are (3, 2) and (2, 3).

18.10 Solve the system: (1) x2 þ 3xy þ 2y2 ¼ 3,

(x  3)(x  2) ¼ 0

and

x ¼ 3, 2:

(2) x2 þ 5xy þ 6y2 ¼ 15.

SOLUTION Dividing (1) by (2), x2 þ 3xy þ 2y2 (x þ y)(x þ 2y) xþy 1 ¼ ¼ ¼ : x2 þ 5xy þ 6y2 (x þ 3y)(x þ 2y) x þ 3y 5 xþy 1 ¼ , y ¼ 2x. Substituting y ¼ 2x in (1) or (2), x2 ¼ 1 and x ¼ +1: x þ 3y 5 The solutions are (1, 22) and (21, 2).

From

18.11 Solve the system: (1) x2 þ y2 þ 2x þ 2y ¼ 32,

(2) x þ y þ 2xy ¼ 22:

SOLUTION The equations are symmetric in x and y since interchange of x and y yields the same equation. Substituting x ¼ u þ v, y ¼ u 2 v in (1) and (2), we obtain (3) u2 þ v2 þ 2u ¼ 16

and

(4) u2  v2 þ u ¼ 11:

Adding (3) and (4), we get 2u 2 þ 3u 2 27 ¼ 0, (u 2 3)(2u þ 9) ¼ 0 and u ¼ 3, 29/2.

196

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

[CHAP. 18

pffiffiffiffiffi When u ¼ 3, v2 ¼ 1 and v ¼ +1; when u ¼ 29/2, v2 ¼ 19/4 and v ¼ + p 19ffiffiffiffiffi =2. Thus the solutions of (3) and pffiffiffiffiffi (4) are: u ¼ 3, v ¼ 1; u ¼ 3, v ¼ 1; u ¼ 9=2, v ¼ 19=2; u ¼ 9=2, v ¼  19=2. Then, since x ¼ u þ v, y ¼ u 2 v, the four solutions of (1) and (2) are: (4, 2);

(2, 4);

pffiffiffiffiffi pffiffiffiffiffi  9 þ 19 9  19 , ; 2 2



pffiffiffiffiffi pffiffiffiffiffi 9  19 9 þ 19 , : 2 2

18.12 Solve the system: (1) x2 þ y2 ¼ 180,

(2)

1 1 1 þ ¼ : x y 4

SOLUTION From (2) obtain (3) 4x þ 4y 2 xy ¼ 0. Since (1) and (3) are symmetric in x and y, substitute x ¼ u þ v, y ¼ u 2 v in (1) and (3) and obtain (4) u2 þ v2 ¼ 90

and

(5) 8u  u2 þ v2 ¼ 0:

0,ffiffiffiffiffi (u 2 9)(u þ 5) ¼ 0 and u ¼ 9, 25. Subtracting (5) from (4), we have u 2 2 4u 2 45 ¼ p When u ¼ 9, v ¼ +3; when u ¼p5, ffiffiffiffiffi v ¼ + 65. Thus pffiffiffiffiffi the solutions of (4) and u ¼ 9, v ¼ 3; u ¼ 9, v ¼ 3; u ¼ 5, v ¼ 65; u ¼ 5, v ¼  65. Hence the four solutions of (1) and (2) are: pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi (12, 6); (6, 12); (5 þ 65, 5  65); (5  65, 5 þ 65):

(5)

are:

18.13 The sum of two numbers is 25 and their product is 144. What are the numbers? SOLUTION Let the numbers be x, y. Then (1) x þ y ¼ 25 and (2) xy ¼ 144. The simultaneous solutions of (1) and (2) are x ¼ 9, y ¼ 16 and x ¼ 16, y ¼ 9. Hence the required numbers are 9, 16.

18.14 The difference of two positive numbers is 3 and the sum of their squares is 65. Find the numbers. SOLUTION Let the numbers be p, q. Then (1) p 2 q ¼ 3 and (2) p 2 þ q 2 ¼ 65. The simultaneous solutions of (1) and (2) are p ¼ 7, q ¼ 4 and p ¼ 24, q ¼ 27. Hence the required (positive) numbers are 7, 4.

18.15 A rectangle has perimeter 60 ft and area 216 ft2. Find its dimensions. SOLUTION Let the rectangle have sides of lengths x, y. Then (1) 2x þ 2y ¼ 60 and (2) xy ¼ 216. Solving (1) and (2) simultaneously, the required sides are 12 and 18 ft.

18.16 The hypotenuse of a right triangle is 41 ft long and the area of the triangle is 180 ft2. Find the lengths of the two legs. SOLUTION Let the legs have lengths x, y. Then (1) x 2 þ y 2 ¼ (41)2 and (2) 12(xy) ¼ 180. Solving (1) and (2) simultaneously, we find the legs have lengths 9 and 40 ft.

CHAP. 18]

197

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

Supplementary Problems 18.17

Solve the following systems graphically. (a) x2 þ y2 ¼ 20, 3x  y ¼ 2 (b) x2 þ 4y2 ¼ 25, x2  y2 ¼ 5

18.18

(c) y2 ¼ x, x2 þ 2y2 ¼ 24 (d ) x2 þ 1 ¼ 4y, 3x  2y ¼ 2

Solve the following systems algebraically. (a) 2x2  y2 ¼ 14, x  y ¼ 1

(h) x2 þ 3xy ¼ 18, x2  5y2 ¼ 4

(b) xy þ x2 ¼ 24, y  3x þ 4 ¼ 0

(i) x2 þ 2xy ¼ 16, 3x2  4xy þ 2y2 ¼ 6

(c) 3xy  10x ¼ y, 2  y þ x ¼ 0

( j) x2  xy þ y2 ¼ 7, x2 þ y2 ¼ 10

(d ) 4x þ 5y ¼ 6, xy ¼ 2

(k) x2  3y2 þ 10y ¼ 19, x2  3y2 þ 5x ¼ 9

(e) 2x  y ¼ 5, 3x þ 4y ¼ 57

(l) x3  y3 ¼ 9, x  y ¼ 3

( f ) 9=x2 þ 16=y2 ¼ 5, 18=x2  12=y2 ¼ 1

(m) x3  y3 ¼ 19, x2 y  xy2 ¼ 6

(g) x  xy ¼ 12, xy  y ¼ 3

(n) 1=x3 þ 1=y3 ¼ 35, 1=x2  1=xy þ 1=y2 ¼ 7

2

2

2

2

2

2

18.19

The square of a certain number exceeds twice the square of another number by 16. Find the numbers if the sum of their squares is 208.

18.20

The diagonal of a rectangle is 85 ft. If the short side is increased by 11 ft and the long side decreased by 7 ft, the length of the diagonal remains the same. Find the dimensions of the original rectangle.

ANSWERS TO SUPPLEMENTARY PROBLEMS 18.17

(a) (2, 4), (20.8, 24.4) See Fig. 18-2 (b) (3, 2), (23, 2), (3, 22), (23, 22) See Fig. 18-3

18.18

(a) (b) (c) (d ) (e) (f ) (g) (h)

18.19

12, 8; 212, 28; 12, 28; 212, 8

18.20

40 ft, 75 ft

(c) (4, 2), (4, 22) See Fig. 18-4 (d) (1, 0.5), (5, 6.5) See Fig. 18-5

(3, 2), (25, 26) (i) (3, 5), (22, 210) ( j) (2, 4), (21/3, 5/3) (k) (21, 2), (5/2, 24/5) (l ) pffiffiffi pffiffiffi pffiffiffi pffiffiffi ( 7, 3), ( 7, 3), ( 7, 3), ( 7, 3) (m) (3, 2), (3, 22), (23, 2), (23, 22) (n) pffiffiffi  pffiffiffi (4, 1), (24, 21)  pffiffiffi pffiffiffi 7i 5 7i 5 (3, 1), (23, 21), 3i 5, , 3i 5, 5 5

(2, 3), (22, 23) (1, 3), (21, 23), (3, 1), (23, 21) (212, 25), (4, 3) (1, 22), (2, 21) (22, 23), (3, 2) (1/2, 1/3), (1/3, 1/2)

(–3,2)

(3,2)

(–3,–2) 8

Fig. 18-2

Fig. 18-3

(3,–2)

198

SYSTEMS OF EQUATIONS INVOLVING QUADRATICS

[CHAP. 18

x −

O − −

Fig. 18-4

x −

O −

Fig. 18-5

CHAPTER 19

Inequalities 19.1

DEFINITIONS

An inequality is a statement that one real quantity or expression is greater or less than another real quantity or expression. The following indicate the meaning of inequality signs. (1) a . b means “a is greater than b” (or a  b is a positive number). (2) a , b means “a is less than b” (or a  b is a negative number). (3) a  b means “a is greater than or equal to b.” (4) a  b means “a is less than or equal to b.” (5) 0 , a , 2 means “a is greater than zero but less than 2.” (6) 2  x , 2 means “x is greater than or equal to 22 but less than 2.” An absolute inequality is true for all real values of the letters involved. For example, (a  b)2 . 1 holds for all real values of a and b, since the square of any real number is positive or zero. A conditional inequality holds only for particular values of the letters involved. Thus x  5 . 3 is true only when x is greater than 8. The inequalities a . b and c . d have the same sense. The inequalities a . b and x , y have opposite sense. 19.2

PRINCIPLES OF INEQUALITIES

(1) The sense of an inequality is unchanged if each side is increased or decreased by the same real number. It follows that any term may be transposed from one side of an inequality to the other, provided the sign of the term is changed. Thus if a . b, then a þ c . b þ c, and a 2 c . b 2 c, and a 2 b . 0. (2) The sense of an inequality is unchanged if each side is multiplied or divided by the same positive number. Thus if a . b and k . 0, then a b ka . kb and . : k k (3) The sense of an inequality is reversed if each side is multiplied or divided by the same negative number. Thus if a . b and k , 0, then a b ka , kb and , : k k (4) If a . b and a, b, n are positive, then a n . b n but an , bn . 199

200

INEQUALITIES

[CHAP. 19

EXAMPLES 19.1. 5 . 4; then 53 . 43

or

16 . 9; then 161=2 . 91=2

125 . 64, but 53 , 43 or

or

4 . 3, but 161=2 , 91=2

1 1 , . 125 64 1 1 or , . 4 3

(5) If a . b and c . d, then (a þ c) . (b þ d ). (6) If a . b . 0 and c . d . 0, then ac . bd.

19.3

ABSOLUTE VALUE INEQUALITIES

The absolute value of a quantity represents the distance that the value of the expression is from zero on a number line. So jx  aj ¼ b, where b . 0, says that the quantity x  a is b units from 0, x  a is b units to the right of 0, or x  a is b units to the left of 0. When we say jx  aj . b, b . 0, then x  a is at a distance from 0 that is greater than b. Thus, x  a . b or x  a , b. Similarly, if jx  aj , b, b . 0, then x  a is at a distance from 0 that is less than b. Hence, x  a is between b units below 0, b, and b units above 0. EXAMPLES 19.2. (a) jx  3j . 4

Solve each of these inequalities for x. (b) jx þ 4j , 7

(c) jx  5j , 3

(d ) jx  5j . 5

(a) jx  3j . 4, then x  3 . 4 or x  3 , 4. Thus, x . 7 or x , 1. The solution interval is (1, 1) < (7, 1), (where < represents the union of the two intervals). (b) jx þ 4j , 7 then 7 , x þ 4 , 7. Thus, 11 , x , 3. The solution interval is (11, 3). (c) jx  5j , 3 Since the absolute value of a number is always greater than or equal to zero, there are no values for which the absolute value will be less than 23. Thus, there is no solution and we may write 1 for the solution interval. (d ) jx þ 3j . 5 Since the absolute value of a number is always at least zero, it is always greater than 5. Thus the solution is all real numbers, and for the solution interval we write (1, 1).

19.4

HIGHER DEGREE INEQUALITIES

Solving higher degree inequalities is similar to solving higher degree equations: we must always compare the expression to zero. If f (x) . 0, then we are interested in the values of x that will produce a product and/ or quotient of factors that is positive, while if f (x) , 0, we wish to find the values of x that will produce a product and/or quotient that is negative. If f(x) is a quadratic expression we have just two factors to consider, and we can do this by examining cases based on the possible signs of the two factors that will produce the desired sign for the expression (see Problems 19.3(c) and 19.14). When the number of factors in f(x) increases by one, the number of cases to consider doubles. Thus, for an expression with 2 factors there are 4 cases, with 3 factors there are 8 cases, and with 4 factors there are 16 cases. In each instance half of the cases will produce a positive expression and half a negative one. Thus, the case procedure gets to be a very long one quite quickly. An alternative procedure to the case method is the sign chart. EXAMPLE 19.3. Solve the inequality x2 þ 15 , 8x. The inequality x2 þ 15 , 8x is equivalent to x2  8x þ 15 , 0 and to (x  3)(x  5) , 0 and is true when the product of x  3 and x  5 is negative. The critical values of the product are the values that make these factors 0, because they represent where the product may change signs. The critical values of x, 3 and 5, are placed on a number line and divide it into three intervals. We need to find the sign of the product of x  3 and x  5 on each of these intervals to find the solution (see Fig. 19-1). Vertical lines are drawn through each critical value. A dashed line indicates that the critical value is not in the solution and a solid line indicates that the critical value is in the solution.

CHAP. 19]

INEQUALITIES

201

The signs above the number line are the signs for the factors and are found by selecting an arbitrary value in the interval as a test value and determining whether each factor is positive or negative for the test value. For the interval to the left of 3, we choose a test value of 1 and substitute it into x  3 and see that the value is 2, so we record a 2 sign, and for x  5 the value is 4 and again we record a 2 sign. For the interval between 3 and 5 we choose any value, such as 3.5, and determine that x  3 is positive and x  5 is negative. Finally, for the interval to the right of 5, we choose a value of 12 and see that both x  3 and x  5 are positive. The sign for the problem, written below the line, in each interval is determined by the signs of the factors in that interval. If an even number of factors in a product or quotient are negative, the product or quotient is positive. If an odd number of factors are negative, the product or quotient is negative.

Fig. 19-1 We select the intervals that satisfy our problem (x  3)(x  5) , 0, so we select the intervals that are negative in the sign chart. In the interval between 3 and 5 the problem is negative (see Fig. 19-1), so the solution is the interval (3, 5). The parentheses mean that the 3 and 5 are not included in the interval, and we know this since the boundary lines are dashed. If they had been in the solution, we would have used a bracket instead of a parenthesis at the end of the interval next to the 3. The solution for x2 þ 15 , 8x is the interval (3, 5). EXAMPLE 19.4.

Solve the inequality x3  0: x(x þ 4)

The inequality is compared to 0 and the numerator and denominator are factored, so we can see that the critical values for the problem are the solution of x ¼ 0, x  3 ¼ 0, and x þ 4 ¼ 0. Thus, the critical values are x ¼ 0, x ¼ 3 and x ¼ 4. Since there are three critical values, the number line is divided into four distinct intervals, as shown in Fig. 19-2.

Fig. 19-2 The signs above the line are the signs of each factor in each interval. The sign below is the sign for the problem and it is þ when an even number of factors are negative and 2 when an odd number of factors are negative. Since the problem uses the  sign, values that make the numerator zero are solutions, so a solid line is drawn through 3. Since 0 and 4 make the denominator of the fraction 0, they are not solutions and dashed lines were drawn through 0 and 4 (see Fig. 19-2). Since the problem x3 0 x(x þ 4) indicates that a positive or zero value is wanted, we want the regions with a þ sign in the sign chart. Thus, the solutions are the intervals, (4, 0) and ½3, 1), and the solution is written (4, 0) < ½3, 1). The < indicates that we want the union of the two intervals. Note that the bracket, [, is used because the critical value 3 is in the solution and a parenthesis, ), is always used for the infinite, 1, side of an interval.

202

INEQUALITIES

19.5

LINEAR INEQUALITIES IN TWO VARIABLES

[CHAP. 19

The solution of linear inequalities in two variables x and y consists of all points (x, y) that satisfy the inequality. Since a linear equation represents a line, a linear inequality is the points on one side of a line. The points on the line are included when the sign  or  is used in the statement of the inequality. The solutions of linear inequalities are usually found by graphical methods. EXAMPLE 19.5. Find the solution for 2x  y  3. We graph the line related to the inequality 2x  y  3, which is 2x  y ¼ 3. Since the symbol  is used, the line is part of the solution and a solid line is used to indicate this (see Fig. 19-3). If the line is not part of the solution, we use a dashed line to indicate that fact. We shade the region on the side of the line where the points are solutions of the inequality. The solution region is determined by selecting a test point that is not on the line. If the test point satisfies the inequality, then all points on that side of the line are in the solution. If the test point does not satisfy the inequality, no points on that side of the line are in the solution. Hence the solution points are on the opposite side of the line from the test point. The point P(2, 4) is not on the line 2x  y ¼ 3, so it can be used as a test point. When we substitute (2, 4) into the inequality 2x  y  3, we get 2(2)  4  3, which is true, since 0  3: We shade on the side of the line that contains the test point (2, 4) to indicate the solution region. If we had selected Q(5, 2) and substituted into 2x  y  3, we would have obtained 12  3, which is false, and would have shaded on the opposite side of the line from Q. This is the same region we found using the test point P. The solution for 2x  y  3 is shown in Fig. 19-3 and consists of the shaded region and the line.

Fig. 19-3

19.6

Fig. 19-4

SYSTEMS OF LINEAR INEQUALITIES

If we have two or more linear inequalities in two variables, we say we have a system of linear inequalities and the solution of the system is the intersection, or common region, of the solution regions for the inequalities. A system with two inequalities whose related equations intersect always has a solution region. If the related equations are parallel, the system may or may not have a solution. Systems with three or more inequalities may or may not have a solution. EXAMPLE 19.6.

Solve the system of inequalities 2x þ y . 3 and x  2y  1.

CHAP. 19]

INEQUALITIES

203

We graph the related equations 2x þ y ¼ 3 and x  2y ¼ 1 on the same set of axes. The line 2x þ y ¼ 3 is dashed, since it is not included in 2x þ y . 3, but the line x  2y ¼ 1 is solid, since it is included in x  2y  1. Now we select a test such as (0, 5) that is not on either line, determine which side of each line to shade and shade only the common region. Since 2(0) þ 5 . 3 is true, the solution region is to the right and above the line 2x þ y ¼ 3. Since 0  2(5)  1 is true, the solution region is to the left and above the line x  2y ¼ 1. The solution region of 2x þ y . 3 and x  2y  1 is the shaded region of Fig. 19-4, which includes the part of the solid line bordering the shaded region.

19.7

LINEAR PROGRAMMING

Many practical problems from business involve a function (objective) that is to be either maximized or minimized subject to a set of conditions (constraints). If the objective is a linear function and the constraints are linear inequalities, the values, if any, that maximize or minimize the objective occur at the corners of the region determined by the constraints. EXAMPLE 19.7. The Green Company uses three grades of recycled paper, called grades A, B, and C, produced from scrap paper it collects. Companies that produce these grades of recycled paper do so as the result of a single operation, so the proportion of each grade of paper is fixed for each company. The Ecology Company process produces 1 unit of grade A, 2 units of grade B, and 3 units of grade C for each ton of paper processed and charges $300 for the processing. The Environment Company process produces 1 unit of grade A, 5 units of grade B, and 1 unit of grade C for each ton of paper processed and charges $500 for processing. The Green Company needs at least 100 units of grade A paper, 260 units of grade B paper, and 180 units of grade C paper. How should the company place its order so that costs are minimized? If x represents the number of tons of paper to be recycled by the Ecology Company and y represents the number of tons of paper to be processed by the Environment Company, then the objective function is C(x, y) ¼ 300x þ 500y, and we want to minimize C(x, y). The constraints stated in terms of x and y are for grade A: 1x þ 1y  100; for grade B: 2x þ 5y  260; and for grade C: 3x þ 1y  180. Since you can not have a company process a negative number of tons of paper, x  0 and y  0. These last two constraints are called natural or implied constraints, because these conditions are true as a matter of fact and need not be stated in the problem. We graph the inequalities determined from the constraints (see Fig. 19-5). The vertices of the region are A(0, 180), B(40, 60), C(80, 20), and D(130, 0).

Fig. 19-5

204

INEQUALITIES

[CHAP. 19

The minimum for C(x, y), if it exists, will occur at point A, B, C, or D, so we evaluate the objective function at these points. C(0, 180) ¼ 300(0) þ 500(180) ¼ 0 þ 90 000 ¼ 90 000 C(40, 60) ¼ 300(40) þ 500(60) ¼ 12 000 þ 30 000 ¼ 42 000 C(80, 20) ¼ 300(80) þ 500(20) ¼ 24 000 þ 10 000 ¼ 34 000 C(130, 0) ¼ 300(130) þ 500(0) ¼ 39 000 þ 0 ¼ 39 000 The Green Printing Company can minimize the cost of recycled paper to $34 000 by having the Ecology Company process 80 tons of paper and the Environment Company process 20 tons of paper.

Solved Problems 19.1

If a . b and c . d, prove that a þ c . b þ d. SOLUTION Since (a  b) and (c  d ) are positive, (a  b) þ (c  d ) is positive. Hence (a  b) þ (c  d ) . 0, (a þ c) (b þ d ) . 0 and (a þ c) . (b þ d ).

19.2

Find the fallacy. (a) (b) (c) (d) (e) (f)

Let a ¼ 3, b ¼ 5; then Multiply by a: Subtract b2 : Factor: Divide by a  b: Substitute a ¼ 3, b ¼ 5:

a,b a2 , ab a2  b2 , ab  b2 (a þ b)(a  b) , b(a  b) aþb,b 8,5

SOLUTION There is nothing wrong with steps (a), (b), (c), (d). The error is made in step (e) where the inequality is divided by a – b, a negative number, without reversing the inequality sign.

19.3

Find the values of x for which each of the following inequalities holds. ðaÞ 4x þ 5 . 2x þ 9. We have 4x  2x . 9  5, 2x . 4 and x . 2. (b)

x 1 2x 1  , þ . Multiplying by 6, we obtain 2 3 3 2 3x  2 , 4x þ 3, 3x  4x , 2 þ 3, x , 5, x . 5:

(c) x2 , 16. Method 1. x2  16 , 0, (x  4)(x þ 4) , 0. The product of the factors (x  4) and (x þ 4) is negative. Two cases are possible. (1) x  4 . 0 and x þ 4 , 0 simultaneously. Thus x . 4 and x , 4. This is impossible, as x cannot be both greater than 4 and less than 4 simultaneously. (2) x  4 , 0 and x þ 4 . 0 simultaneously. Thus x , 4 and x . 4. This is possible if and only if 4 , x , 4. Hence 4 , x , 4. Method 2. (x2 )1=2 , (16)1=2 . Now (x2 )1=2 ¼ x if x  0, and (x2 )1=2 ¼ x if x  0. If x  0, (x2 )1=2 , (16)1=2 may be written x , 4. Hence 0  x , 4.

CHAP. 19]

205

INEQUALITIES

If x  0, (x2 )1=2 , (16)1=2 may be written x , 4 or x . 4. Hence 4 , x  0. Thus 0  x , 4 and 4 , x  0, or 4 , x , 4.

19.4

Prove that a2 þ b2 . 2ab if a and b are real and unequal numbers. SOLUTION If a2 þ b2 . 2ab, then a2  2ab þ b2 . 0 or (a  b)2 . 0. This last statement is true since the square of any real number different from zero is positive. The above provides a clue as to the method of proof. Starting with (a  b)2 . 0, which we know to be true if a = b, we obtain a2  2ab þ b2 . 0 or a2 þ b2 . 2ab. Note that the proof is essentially a reversal of the steps in the first paragraph.

19.5

Prove that the sum of any positive number and its reciprocal is never less than 2. SOLUTION We must prove that (a þ 1=a)  2 if a . 0. If (a þ 1=a)  2, then a2 þ 1  2a, a2  2a þ 1  0, and (a  1)2  0 which is true. To prove the theorem we start with (a  1)2  0, which is known to be true. Then a2  2a þ 1  0, a2 þ 1  2a and a þ 1=a  2 upon division by a.

19.6

Show that a2 þ b2 þ c2 . ab þ bc þ ca for all real values of a, b, c unless a ¼ b ¼ c. SOLUTION Since a2 þ b2 . 2ab, b2 þ c2 . 2bc, c2 þ a2 . 2ca (see Problem 19.4), we have by addition or 2(a2 þ b2 þ c2 ) . 2(ab þ bc þ ca) (If a ¼ b ¼ c, then a2 þ b2 þ c2 ¼ ab þ bc þ ca.)

19.7

a2 þ b2 þ c2 . ab þ bc þ ca:

If a2 þ b2 ¼ 1 and c2 þ d2 ¼ 1, show that ac þ bd , 1. SOLUTION a2 þ c2 . 2ac and b2 þ d 2 . 2bd; hence by addition (a2 þ b2 ) þ (c2 þ d 2 ) . 2ac þ 2bd

19.8

or

2 . 2ac þ 2bd, i.e., 1 . ac þ bd:

Prove that x3 þ y3 . x2 y þ y2 x, if x and y are real, positive and unequal numbers. SOLUTION If x3 þ y3 . x2 y þ y2 x, then (x þ y)(x2  xy þ y2 ) . xy(x þ y). Dividing by x þ y, which is positive. x2  xy þ y2 . xy or x2  2xy þ y2 . 0, i.e., (x  y)2 . 0 which is true if x = y: The steps are reversible and supply the proof. Starting with (x  y)2 . 0, x = y, obtain x2  xy þ y2 . xy: Multiplying both sides by x þ y, we have (x þ y)(x2  xy þ y2 ) . xy(x þ y) or x3 þ y3 . x2 y þ y2 x.

19.9

Prove that an þ bn . an1 b þ abn1 , provided a and b are positive and unequal, and n . l. SOLUTION If an þ bn . an1 b þ abn1 , then (an  an1 b)  (abn1  bn ) . 0 an1 (a  b)  bn1 (a  b) . 0,

or

i.e., (an1  bn1 )(a  b) . 0.

206

INEQUALITIES

[CHAP. 19

This is true since the factors are both positive or both negative. Reversing the steps, which are reversible, provides the proof.

19.10 Prove that a3 þ

1 1 . a2 þ 2 a3 a

a.0

if

and

a=1:

SOLUTION Multiplying both sides of the inequality by a3 (which is positive since a . 0), we have a6  a5  a þ 1 . 0 and (a5  1)(a  1) . 0: a6 þ 1 . a5 þ a, If a . 1 both factors are positive, while if 0 , a , 1 both factors are negative. In either case the product is positive. (If a ¼ 1 the product is zero.) Reversal of the steps provides the proof.

19.11 If a, b, c, d are positive numbers and a c . , b d prove that aþc c . : bþd d SOLUTION Method 1. If aþc c . , bþd d then multiplying by d(b þ d ) we obtain (a þ c)d . c(b þ d), ad þ cd . bc þ cd, ad . bc and, dividing by bd, a c . ; b d which is given as true. Reversing the steps provides the proof. Method 2. Since a c . , b d then a c c c þ . þ , b b d b

19.12 Prove: (a) x2  y2 . x  y (b) x2  y2 , x  y

if if

xþy.1 xþy.1

a þ c c(b þ d ) . b bd

and and

and

aþc c . : bþd d

x.y x,y

SOLUTION (a) Since x . y, x  y . 0. Multiplying both sides of x þ y . 1 by the positive number x  y, (x þ y)(x  y) . (x  y)

or

x2  y2 . x  y:

CHAP. 19]

207

INEQUALITIES

(b) Since x , y, x  y , 0. Multiplying both sides of x þ y . 1 by the negative number x  y reverses the sense of the inequality; thus (x þ y)(x  y) , (x  y)

x2  y2 , x  y:

or

19.13 The arithmetic mean of two numbers a and b is (a þ b)/2, the geometric mean is mean is 2ab/(a þ b). Prove that a þ b pffiffiffiffiffi 2ab . ab . 2 aþb

pffiffiffiffiffi ab, and the harmonic

if a and b are positive and unequal. SOLUTION

pffiffiffiffiffi pffiffiffiffiffi 2 2 2 (a) If (a þ b)=2 . ab, then (a þ b)2 . (2 ab)2 , a2 þ 2ab þ b2 . p 4ab, ffiffiffiffiffi a  2ab þ b . 0 and (a  b) . 0 which is true if a = b. Reversing the steps, we have (a þ b)=2 . ab. (b) If pffiffiffiffiffi 2ab ab . , aþb then ab .

4a2 b2 , (a þ b)2

(a þ b)2 . 4ab

which is true if a = b. Reversing the steps, we have From (a) and (b),

and

(a  b)2 . 0

pffiffiffiffiffi ab . 2ab=(a þ b).

a þ b pffiffiffiffiffi 2ab . ab . : 2 aþb

19.14 Find the values of x for which (a) x2  7x þ 12 ¼ 0, (b) x2  7x þ 12 . 0, (c) x2  7x þ 12 , 0. SOLUTION (a) x2  7x þ 12 ¼ (x  3)(x  4) ¼ 0 when x ¼ 3 or 4. (b) x2  7x þ 12 . 0 or (x  3)(x  4) . 0 when (x  3) . 0 and (x  4) . 0 simultaneously, or when (x  3) , 0 and (x  4) , 0 simultaneously. (x  3) . 0 and (x  4) . 0 simultaneously when x . 3 and x . 4, i.e., when x . 4: (x 23) , 0 and (x  4) , 0 simultaneously when x , 3 and x , 4, i.e., when x , 3. Hence x2  7x þ 12 . 0 is satisfied when x . 4 or x , 3. (c) x2  7x þ 12 , 0 or (x  3)(x  4) , 0 when (x  3) . 0 and (x  4) , 0 simultaneously, or when (x  3) , 0 and (x  4) . 0 simultaneously. (x  3) . 0 and (x  4) , 0 simultaneously when x . 3 and x , 4, i.e., when 3 , x , 4: (x  3) , 0 and (x  4) . 0 simultaneously when x , 3 and x . 4, which is absurd. Hence x2  7x þ 12 , 0 is satisfied when 3 , x , 4.

19.15 Determine graphically the range of values of x defined by (a) x2 þ 2x  3 ¼ 0 (b) x2 þ 2x  3 . 0 (c) x2 þ 2x  3 , 0: SOLUTION Figure 19-6 shows the graph of the function defined by y ¼ x2 þ 2x  3. From the graph it is clear that

208

INEQUALITIES

Fig. 19-6 (a) y ¼ 0 when x ¼ 1, x ¼ 3 (b) y . 0 when x . 1 or x , 3 (c) y , 0 when 3 , x , 1.

19.16 Solve for x: (a) j3x  6j þ 2 . 9

(b) j7x  1j  6 , 2.

(a) j3x  6j þ 2 . 9 j3x  6j . 7 3x  6 . 7 or 3x  6 , 7 3x . 13 or 3x , 1 x . 13=3 or x , 1=3 The solution of j3x  6j þ 2 . 9 is the interval (1, 1=3) < (13=3, 1). (b) j7x  1j  6 , 2 j7x  1j , 8 8 , 7x  1 , 8 7 , 7x , 9 1 , x , 9=7 The solution of j7x  1j 6 , 2 is the interval (21, 9/7). 19.17 Solve for x: (a)

2x  1 1 xþ1

(b)

x2  10x þ 21  0: x2  5x þ 6

SOLUTION (a)

2x  1 1 xþ1

2x  1 10 xþ1 2x  1 x þ 1  0 xþ1 xþ1 x2 0 xþ1

[CHAP. 19

CHAP. 19]

209

INEQUALITIES

The critical values are x ¼ 1 and x ¼ 2. We make a sign chart (see Fig. 19-7), with a solid line through x ¼ 2, since it makes the fraction 0, and 0 is included in the solution, and a dashed line through x ¼ 1, since it makes the fraction undefined. Next, we determine the sign of each factor in the three intervals. Finally, in intervals where an even number of factors are negative the problem is positive and in those where an odd number of factors are negative the problem is negative. The solution of 2x  1 1 xþ1 is the interval (1, 2. (b)

x2  10x þ 21 0 x2  5x þ 6 (x  3)(x  7) 0 (x  3)(x  2) The critical values are x ¼ 2, x ¼ 3, and x ¼ 7. We make a sign chart (see Fig. 19-8), with dashed lines through x ¼ 2 and x ¼ 3 and a solid line through x ¼ 7. Since x ¼ 3 makes the denominator of the fraction zero it is excluded, even though it also makes the numerator 0. The signs for the factors are determined for each interval and then used to determine the sign for the problem in each interval. The factor x  3 is used an even number of times in the problem and could be omitted from the sign chart, since any factor raised to an even power is always non-negative. The solution of x2  10x þ 21 0 x2  5x þ 6 is the interval (2, 3) < (3, 7]. Note 1: If we had canceled the common factor x  3, we might have overlooked the fact that the problem is not defined when x ¼ 3 and it cannot be in the solution set. Note 2: When a factor appears in the problem an even number of times, it may be excluded from the sign chart and is usually omitted. When a factor appears in the problem an odd number of times it must be included in the sign chart an odd number of times, and is usually included exactly once.

Fig. 19-7

Fig. 19-8

19.18 Find the solution for the system of inequalities 2x þ y  2 and 2x  y  6. SOLUTION Graph the related equations 2x þ y ¼ 2 and 2x  y ¼ 6. Both lines are solid, since they are included in the solution. Using (0, 0) as the test point, we get 2(0) þ 0  2, which is false, and 2(0)  0  6, which is true. Since the test point (0, 0) makes 2x þ y  2 false, the solution lies on the opposite side of the line 2x þ y ¼ 2 from the point (0, 0). So we shade above and to the left of the line 2x þ y ¼ 2. Since the test point (0, 0) makes 2x  y  6 true, the solution is on the same side of the line 2x  y ¼ 6 as the point (0, 0). So we shade above and to the left of the line 2x  y ¼ 6. The common solution is the region above and to the left of 2x þ y ¼ 2, and is the shaded region shown in Fig. 19-9.

210

INEQUALITIES

[CHAP. 19

Fig. 19-9

19.19 The Close Shave Company manufactures two types of electric shaver. One shaver is cordless, requires 4 hours to make, and sells for $40. The other shaver is a cord type, takes 2 hours to make, and sells for $30. The Company has only 800 work hours to use in manufacturing each day and the shipping department can pack and ship only 300 shavers per day. How many of each type of shaver should the Close Shave Company produce per day to maximize its sales revenue? SOLUTION Let x be the number of cordless shavers made per day and y be the number of cord-type shavers made per day. The objective function is R(x, y) ¼ 40x þ 30y. The stated constraints are 4x þ 2y  800 and x þ y  300. The natural constraints are x  0 and y  0. From Fig. 19-10, we see that the vertices of the region formed by the constraints are A(0, 0), B(200, 0), C(100, 200), and D(0, 300). R(0, 0) ¼ 40(0) þ 30(0) ¼ 0 þ 0 ¼ 0 R(200, 0) ¼ 40(200) þ 30(0) ¼ 8000 þ 0 ¼ 8000 R(100, 200) ¼ 40(100) þ 30(200) ¼ 4000 þ 6000 ¼ 10 000 R(0, 300) ¼ 40(0) þ 30(300) ¼ 0 þ 9000 ¼ 9000: The Close Shave Company achieves the maximum sales revenue of $10 000 per day by producing 100 cordless shavers and 200 cord-type shavers per day.

Supplementary Problems 19.20

If a . b, prove that a  c . b  c where c is any real number.

19.21

If a . b and k . 0, prove that ka . kb.

19.22

Find the values of x for which of the following inequalities holds. (a) 2(x þ 3) . 3(x  1) þ 6

(b)

x 2 2x 1 þ ,  4 3 3 6

(c)

1 3 7 þ . x 4x 8

(d ) x2 . 9

19.23

For what values of a will (a þ 3) , 2(2a þ 1)?

19.24

Prove that 12 (a2 þ b2 )  ab for all real values of a and b, the equality holding if and only if a ¼ b.

CHAP. 19]

211

INEQUALITIES

Fig. 19-10 19.25

Prove that 1 1 2 þ . x y xþy if x and y are positive and x = y.

19.26

Prove that x2 þ y2 ,xþy xþy

if

x . 0, y . 0:

19.27

Prove that xy þ 1  x þ y if x  1 and y  1 or if x  1 and y  1.

19.28

If a . 0, a = 1 and n is any positive integer, prove that anþ1 þ pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 þ 6 , 3 þ 5.

1 1 . an þ n : anþ1 a

19.29

Show that

19.30

Determine the values of x for which each of the following inequalities holds. (a) x2 þ 2x  24 . 0

(b) x2  6 , x

(c) 3x2  2x , 1

(d) 3x þ

1 7 . x 2

19.31

Determine graphically the range of values of x for which (a) x2  3x  4 . 0, (b) 2x2  5x þ 2 , 0.

19.32

Write the solution for each inequality in interval notation. (a) j3x þ 3j  15  6

19.33

(b) j2x  3j , 7

Write the solution for each inequality in the interval notation. (a) x2  10x  21 (b)

(x þ 1)(x  1) ,0 x

(c) (x  1)(x  2)(x þ 3) . 0 (d )

x1 0 xþ2

(e)

x5 3 xþ1

( f)

(x  6)(x  3) 0 xþ2

212

19.34

INEQUALITIES

Graph each inequality and shade the solution region. (a) 4x  y  5

19.35

(b) y  3x . 2

Graph each system of inequalities and shade the solution region. (a) x þ 2y  20 and 3x þ 10y  80 (b) 3x þ y  4, x þ y  2, x þ y  4,

19.36

[CHAP. 19

and

x5

Use linear programming to solve each problem. (a) Ramone builds portable storage buildings. He uses 10 sheets of plywood and 15 studs in a small building and 15 sheets of plywood and 45 studs in a large building. Ramone has 60 sheets of plywood and 135 studs available for use. If Ramone makes a profit of $400 on a small building and $500 on a large building, how many of each type of building should he make to maximize his profit? (b) Jean and Wesley make wind chimes and bird houses in their craft shop. Each wind chime requires 3 hours of work from Jean and 1 hour of work from Wesley. Each bird house requires 4 hours of work from Jean and 2 hours of work from Wesley. Jean cannot work more than 48 hours per week and Wesley cannot work more than 20 hours per week. If each wind chime sells for $12 and each bird house sells for $20, how many of each item should they make to maximize their revenue?

ANSWERS TO SUPPLEMENTARY PROBLEMS 19.22

(a) x , 3

19.23

a.

19.30

(a) x . 4

19.31

(a) x . 4 or x , 1

19.32

(a) (1, 4) < ½2, 1)

(b) (2, 5)

19.33

(a) (1, 3) < ½7, 1)

(c) (3, 1) < (2, 1)

(e) (1,  4 < (1, 1)

(b) (1, 1) < (0, 1)

(d) (2, 1

( f ) (2, 3) < ½6, 1)

19.34

(b) x . 2

(c) 0 , x , 2

(d) x , 3 or

x.3

1 3 or

x , 6

(a) Figure 19-11

(b) 2 , x , 3 (b)

1 (c)  , x , 1 3

(d ) x .

2 or 3

1 ,x,2 2

(b) Figure 19-12

Fig. 19-11

Fig. 19-12

0,x,

1 2

CHAP. 19]

213

INEQUALITIES

Fig. 19-13

Fig. 19-14

19.35

(a) Figure 19-13

(b) Figure 19-14

19.36

(a) Ramone maximizes his profit by making 6 small buildings and 0 large buildings. (b) Jean and Wesley will maximize their revenue by making 6 wind chimes and 8 bird houses.

CHAPTER 20

Polynomial Functions 20.1

POLYNOMIAL EQUATIONS

A rational integral equation of degree n in the variable x is an equation which can be written in the form an xn þ an1 xn1 þ an2 xn2 þ    þ a1 x þ a0 ¼ 0,

an = 0

a2ffiffi,ffi . . . , an1 , an are constants. where n is a positive integer and a0 , a1 , p pffiffiffiffiffiffiffi Thus 4x3  2x2 þ 3x  5 ¼ 0, x2  2x þ 14 ¼ 0 and x4 þ 3x  8 ¼ 0 are rational integral equations in x of degree 3, 2 and 4 respectively. Note that in each equation the exponents of x are positive and integral, and the coefficients are constants (real or complex numbers). The coefficient of the highest degree term is called the lead coefficient and a0 is called the constant term. In this chapter, only rational integral equations are considered. A polynomial of degree n in the variable x is a function of x which can be written in the form P(x) ¼ an xn þ an1 xn1 þ an2 xn2 þ    þ a1 x þ a0 ,

an = 0

where n is a positive integer and a0 , a1 , a2 , . . . , an1 , an are constants. Then P(x) ¼ 0 is a rational integral equation of degree n in x. 3 2 3(2) þ (2) If P(x) ¼ 3x3 þ x2 þ 5x  6, then pffiffiffi P(2) ¼ p ffiffiffi pffiffiffi þ 5(2)  6 ¼ 36. 2 If P(x) ¼ x þ 2x  8, then P( 5) ¼ 5 þ 2 5  8 ¼ 2 5  3. Any value of x which makes P(x) vanish is called a root of the equation P(x) ¼ 0. Thus 2 is a root of the equation P(x) ¼ 3x3  2x2  5x  6 ¼ 0, since P(2) ¼ 24 2 8 2 10 2 6 ¼ 0.

20.2 A.

ZEROS OF POLYNOMIAL EQUATIONS Remainder theorem If r is any constant and if a polynomial P(x) is divided by (x  r), the remainder is P(r). For example, if P(x) ¼ 2x3  3x2  x þ 8 is divided by x þ 1, then r ¼ 1 and the remainder ¼ P(1) ¼ 2  3 þ 1 þ 8 ¼ 4. That is, 2x3  3x2  x þ 8 4 ¼ Q(x) þ , xþ1 xþ1 214

where Q(x) is a polynomial in x:

CHAP. 20]

POLYNOMIAL FUNCTIONS

215

B.

Factor theorem If r is a root of the equation P(x) ¼ 0, i.e. if P(r) ¼ 0, then (x  r) is a factor of P(x). Conversely, if (x  r) is a factor of P(x), then r is a root of P(x) ¼ 0, or P(r) ¼ 0. Thus 1, 2, 3 are the three roots of the equation P(x) ¼ x3 þ 4x2 þ x  6 ¼ 0, since P(1) ¼ P(2) ¼ P(3) ¼ 0. Then (x  1), (x þ 2) and (x þ 3) are factors of x3 þ 4x2 þ x  6.

C.

Synthetic division Synthetic division is a simplified method of dividing a polynomial P(x) by x  r, where r is any assigned number. By this method we determine values of the coefficients of the quotient and the value of the remainder can readily be determined. EXAMPLE 20.1. Divide (5x þ x4  14x2 ) by (x þ 4) using synthetic division. Write the terms of the dividend in descending powers of the variable and fill in missing terms using zero for the coefficients; write the divisor in the form x  a. (x4 þ 0x3  14x2 þ 5x þ 0) 4 (x  (4)) Write the constant term a from the divisor on the left in a right of the symbol

j and write the coefficients from the dividend to the

4 j 1 þ 0  14 þ 5 þ 0 Bring down the first term in the divisor to the third row, leaving a blank row for now. 4 j 1 þ 0  14 þ 5 þ 0 1 Multiply the term in the quotient row (third row) by the divisor and write the product in the second row under the second term in the first row, add the numbers in the column formed, and write the sum as the second term in the quotient row. 4 j 1 þ 0  14 þ 5 þ 0 4 14 Multiply the last term on the right in the quotient row by the divisor, write it under the next term in the top row, add, and write the sum in the quotient row. Continue this process until all of the terms in the top row have a number under them. 4 j

1 þ 0  14 þ 5 þ 0  4 þ 16  8 þ 12 1  4 þ 2  3 þ 12

The third row is the quotient row with the last term being the remainder. The degree of the quotient polynomial is one less than the degree of the dividend because we are dividing by a linear factor. The terms of the quotient row are the coefficients of the terms in the quotient polynomial. The degree of the quotient polynomial here is 3. The quotient with remainder for (5x þ x4  14x2 ) 4 (x þ 4) is 1x3  4x2 þ 2x  3 þ

D.

12 xþ4

Fundamental theorem of algebra Every polynomial equation P(x) ¼ 0 has at least one root, real or complex. 5 Thus x7  3x pffiffiffiþ 2 ¼ 0 has at least one root. But f (x) ¼ x þ 3 ¼ 0 has no root, since no number r exists such that f(r) ¼ 0. Since this equation is not rational, the fundamental theorem does not apply.

216

E.

20.3

POLYNOMIAL FUNCTIONS

[CHAP. 20

Number of roots of an equation Every rational integral equation P(x) ¼ 0 of the nth degree has exactly n roots. Thus 2x3 þ 5x2  14x  8 ¼ 0 has exactly 3 roots, namely 2, 12, 4. Some of the n roots may be equal. Thus the equation of the sixth degree (x  2)3 (x  5)2 (x þ 4) ¼ 0 has 2 as a triple root, 5 as a double root, and 4 as a single root; i.e., the six roots are 2, 2, 2, 5, 5, 4: SOLVING POLYNOMIAL EQUATIONS

A.

Complex and irrational roots (1) If a complex number a þ bi is a root of the rational integral equation P(x) ¼ 0 with real coefficients, then the conjugate complex number a  bi is also a root. It follows that every rational integral equation of odd degree with real coefficients has at least one real root. pffiffiffi (2) If the rational integral pffiffiffi equation P(x) ¼ 0 with pffiffiffirational coefficients has a þ b as a root, where a and b are rational and b is irrational, then a  b is also a root.

B.

Rational root theorem If b/c, a rational fraction in lowest terms, is a root of the equation an xn þ an1 xn1 þ an2 xn2 þ    þ a1 x þ a0 ¼ 0,

an = 0

with integral coefficients, then b is a factor of a0 and c is a factor of an . Thus if b/c is a rational root of 6x3 þ 5x2  3x  2 ¼ 0, the values of b are limited to the factors of 2, which are +1, +2; and the values of c are limited to the factors of 6, which are +1, +2, +3, +6. Hence the only possible rational roots are +1, +2, +1/2, +1/3, +1/6, +2/3. C.

Integral root theorem It follows that if an equation P(x) ¼ 0 has integral coefficients and the lead coefficient is 1: xn þ an1 xn1 þ an2 xn2 þ    þ a1 x þ a0 ¼ 0, then any rational root of P(x) ¼ 0 is an integer and a factor of a0 . Thus the rational roots, if any, of x3 þ 2x2  11x  12 ¼ 0 are limited to the integral factors of 12, which are +1, +2, +3, +4, +6, +12.

D.

Intermediate value theorem If P(x) ¼ 0 is a polynomial equation with real coefficients, then approximate values of the real roots of P(x) ¼ 0 may be found by obtaining the graph of y ¼ P(x) and determining the values of x at the points where the graph intersects the x-axis (y ¼ 0). Fundamental in this procedure is the fact that if P(a) and P(b) have opposite signs then P(x) ¼ 0 has at least one root between x ¼ a and x ¼ b. This fact is based on the continuity of the graph of y ¼ P(x) when P(x) is a polynomial with real coefficients. EXAMPLE 20.2. For each real zero of P(x) ¼ 2x3  5x2  6x þ 4 isolate the zero between two consecutive integers. Since P(x) ¼ 2x3  5x2  6x þ 4 has degree 3, there are at most 3 real zeros. We will look for the real zeros in the interval 5 to 5. The interval is arbitrary and may need to be expanded if the real zeros are not found here. By synthetic division, we will find the value of P(x) for each integer in the interval selected. The remainders from the synthetic division are the values of P(x) and are summarized in the table below. x

25

24

23

22

21

0

1

2

3

4

5

P(x)

2341

2180

277

220

3

4

25

212

25

28

99

CHAP. 20]

217

POLYNOMIAL FUNCTIONS

Note that P(2) ¼ 20 and P(1) ¼ 3 have opposite signs, so from the Intermediate Value Theorem there is a real zero between 2 and 1. Similarly, since P(0) ¼ 4 and P(1) ¼ 5 there is a real zero between 0 and 1, and since P(3) ¼ 5 and P(4) ¼ 28 there is a real zero between 3 and 4. Three real zeros have been isolated, so we have located all the real zeros of P(x). It is not always possible to locate all the real zeros this way because there could be more than one zero between two consecutive integers. When there is an even number of zeros between two consecutive integers the Intermediate Value Theorem will not reveal them when we use just integers for x. The Intermediate Value Theorem does not tell you how many real zeros are in the interval, just that there is at least one real zero in the interval.

E.

Upper and lower limits for the real roots A number a is called an upper limit or upper bound for the real roots of P(x) ¼ 0 if no root is greater than a. A number b is called a lower limit or lower bound for the real roots of P(x) ¼ 0 if no root is less than b. The following theorem is useful in determining upper and lower limits. Let P(x) ¼ an xn þ an1 xn1 þ an2 xn2 þ    þ a0 ¼ 0, where a0 , a1 , . . . , an are real and an . 0. Then: (1) If upon synthetic division of P(x) by x  a, where a  0, all of the numbers obtained in the third row are positive or zero, then a is an upper limit for all the real roots of P(x) ¼ 0. (2) If upon synthetic division of P(x) by x  b, where b  0, all of the numbers obtained in the third row are alternately positive and negative (or zero), then b is a lower limit for all the real roots of P(x) ¼ 0.

EXAMPLE 20.3. Find an interval that contains all the real zeros of P(x) ¼ 2x3  5x2 þ 6. We will find the integer, b, that is the least upper limit of the real zeros of P(x) and the integer, a, that is the lower limit on the real zeros of P(x). All real zeros will be in the interval [a, b]. To find a and b we use synthetic division on P(x) ¼ 2x3  5x2 þ 6. 1j

25þ0þ6 þ233 233þ3

2j

25þ0þ6 þ424

3j 25þ0þ 6 þ6þ3þ 9

212þ2

2 þ 1 þ 3 þ 15

When we divide using 3, the quotient row is all positive, so 3 is the smallest integer that is an upper limit for the real zeros of P(x). Thus b ¼ 3. 1 j 2  5 þ 0 þ 6 2þ77 27þ71 When we divide using 1, the quotient row alternates in sign, so 1 is the greatest integer that is a lower limit for the real zeros of P(x). Thus, a ¼ 1. The real zeros of P(x) ¼ 2x3  5x2 þ 6 are in the interval (1, 3) or 1 , x , 3: Since P(1) = 0 and P(3) = 0, we used interval notation that indicates that neither endpoint is a zero.

F.

Descartes’ Rule of Signs If the terms of a polynomial P(x) with real coefficients are arranged in order of descending powers of x, a variation of sign occurs when two consecutive terms differ in sign. For example, x3  2x2 þ 3x  12 has 3 variations of sign, and 2x7  6x5  4x4 þ x2  2x þ 4 has 4 variations of sign. Descartes’ Rule of Signs says that the number of positive roots of P(x) ¼ 0 is either equal to the number of variations of sign of P(x) or is less than that number by an even integer. The number of negative roots of P(x) ¼ 0 is either equal to the number of variations of sign of P(x) or is less than that number by an even integer.

218

POLYNOMIAL FUNCTIONS

[CHAP. 20

Thus in P(x) ¼ x9  2x5 þ 2x2  3x þ 12 ¼ 0 there are 4 variations of sign of P(x); hence the number of positive roots of P(x) ¼ 0 is 4, (4  2) or (4  4). Since P(x) ¼ (x)9  2(x)5 þ 2(x)2  3(x) þ 12 ¼ x9 þ 2x5 þ 2x2 þ 3x þ 12 ¼ 0 has one variation of sign, then P(x) ¼ 0 has exactly one negative root. Hence there are 4, 2, or 0 positive roots, 1 negative root, and at least 9 2 (4 þ 1) ¼ 4 nonreal roots. (There are 4, 6, or 8 nonreal roots. Why?)

20.4

APPROXIMATING REAL ZEROS

In solving a polynomial equation P(x) ¼ 0, it is not always possible to find all the zeros by the previous methods. We have been able to determine the irrational and imaginary zeros when we were able to find quadratic factors that we could solve using the quadratic formula. If we cannot find the quadratic factors of P(x) ¼ 0, we will not be able to solve for the imaginary zeros, but we can often find an approximation for some of the real zeros. To approximate a real zero of P(x) ¼ 0, we must first find an interval that contains a real zero of P(x) ¼ 0. We can do this using the Intermediate Value Theorem to locate to numbers a and b such that P(a) and P(b) have opposite signs. We keep using the Intermediate Value Theorem until we have isolated the real zero in an interval small enough that it will be known to the desired degree of accuracy. EXAMPLE 20.4. Find a real zero of x3 þ 3x þ 8 ¼ 0 correct to two decimal places. By Descartes’ Rule of Signs, P(x) ¼ x3 þ 3x þ 8 has no positive real zeros and 1 negative real zero. Using synthetic division, we find P(2) ¼ 6 and P(1) ¼ 4, so by the Intermediate Value Theorem P(x) ¼ x3 þ 3x þ 8 has a real zero between 2 and 1. We now use synthetic division and the Intermediate Value Theorem to determine the tenths interval containing the zero. The results are summarized in the table below.

x

21.0

21.1

21.2

21.3

21.4

21.5

21.6

21.7

21.8

21.9

P(x)

4

3.37

2.67

1.90

1.06

0.13

20.90 22.01 23.23 24.56

22.0 26

We can see that P(1:5) is positive and P(1:6) is negative so the zero is between 1:6 and 1:5. Now we check for the hundredths digit by using synthetic on the interval between 1:6 and 1:5: We do not have to find all the hundredths values, just a sign change between two consecutive values.

x P(x)

21.50

21.51

21.52

0.13

0.03

20.07

We see that P(1:51) is positive and P(1:52) is negative, so by the Intermediate Value Theorem there is a real zero between 1:51 and 1:52. Since the real zero is located between 1:51 and 1:52, we just need to determine whether it rounds off to 1:51 or 1:52. To do this we find P(1:515), which is about 0:02. This value of P(1:515) is negative and P(1:51) is positive, so we know that the zero is between 1:515 and 1:510, and all the numbers in this interval rounded to two decimal places are 1:51. Thus, to two decimal places the only real zero of x3 þ 3x þ 8 ¼ 0 is 1:51.

To use a graphing calculator to approximate the real zeros of a polynomial we graph the function and use the trace and zoom features of the calculator. After we graph the function, we use the trace feature to locate an interval that contains a real zero using the Intermediate Value Theorem. We then use the zoom feature to focus in on this interval. We continue using the trace and zoom feature until we find two x values that round to the desired degree of accuracy and have function values that are opposite in sign.

CHAP. 20]

POLYNOMIAL FUNCTIONS

219

Solved Problems 20.1

Prove the remainder theorem: If a polynomial P(x) is divided by (x  r) the remainder is P(r). SOLUTION In the division of P(x) by (x  r), let Q(x) be the quotient and R, a constant, the remainder. By definition P(x) ¼ (x 2 r)Q(x) þ R, an identity for all values of x. Letting x ¼ r, P(r) ¼ R.

20.2

Determine the remainder R after each of the following divisions. (a) (2x3 þ 3x2  18x  4) 4 (x  2).

R ¼ P(2) ¼ 2(23 ) þ 3(22 )  18(2)  4 ¼ 12

(b) (x4  3x3 þ 5x þ 8) 4 (x þ 1).

R ¼ P(1) ¼ (1)4  3(1)3 þ 5(1) þ 8



 1 3 2 (c) (4x þ 5x  1) 4 x þ : 2 (d) (x3  2x2 þ x  4) 4 x:   8 3 4 2 3 x  x þx 4 (2x  3): (e) 27 9 2 pffiffiffiffiffiffiffi ( f ) (x8  x5  x3 þ 1) 4 (x þ 1):

¼1þ35þ8¼7    3  2 1 1 1 1 þ5  1 ¼  R¼P  ¼4  2 2 2 4 R ¼ P(0) ¼ 4       3 8 3 3 4 3 2 3 3 R¼P ¼  þ  ¼0 2 27 2 9 2 2 2 R ¼ P(i) ¼ (i)8  (i)5  (i)3 þ 1 ¼ i8 þ i5 þ i3 þ 1 ¼ 1 þ i  i þ 1 ¼ 2

20.3

Prove the factor theorem: If r is a root of the equation P(x) ¼ 0, then (x 2 r) is a factor of P(x); and conversely if (x  r) is a factor of P(x), then r is a root of P(x) ¼ 0. SOLUTION In the division of P(x) by (x  r), let Q(x) be the quotient and R, a constant, the remainder. Then P(x) ¼ (x  r)Q(x) þ R or P(x) ¼ (x  r)Q(x) þ P(r) by the remainder theorem. If r is a root of P(x) ¼ 0, then P(r) ¼ 0. Hence P(x) ¼ (x  r)Q(x), or (x  r) is a factor of P(x). Conversely if (x  r) is a factor of P(x), then the remainder in the division of P(x) by (x  r) is zero. Hence P(r) ¼ 0, i.e., r is a root of P(x) ¼ 0.

20.4

Show that (x  3) is a factor of the polynomial P(x) ¼ x4  4x3  7x2 þ 22x þ 24. SOLUTION P(3) ¼ 81  108  63 þ 66 þ 24 ¼ 0: Hence (x  3) is a factor of P(x), 3 is a zero of the polynomial P(x), and 3 is a root of the equation P(x) ¼ 0.

20.5

(a) Is 1 a root of the equation P(x) ¼ x3  7x  6 ¼ 0? (b) Is 2 a root of the equation P( y) ¼ y4  2y2  y þ 7 ¼ 0? (c) Is 2i a root of the equation P(z) ¼ 2z3 þ 3z2 þ 8z þ 12 ¼ 0? SOLUTION (a) P(1) ¼ 1 þ 7  6 ¼ 0. Hence 1 is a root of the equation P(x) ¼ 0, and ½x  (1) ¼ x þ 1 is a factor of the polynomial P(x). (b) P(2) ¼ 16  8  2 þ 7 ¼ 13. Hence 2 is not a root of P( y) ¼ 0, and ( y  2) is not a factor of y4  2y2  y þ 7. (c) P(2i) ¼ 2(2i)3 þ 3(2i)2 þ 8(2i) þ 12 ¼ 16i  12 þ 16i þ 12 ¼ 0. Hence 2i is a root of P(z) ¼ 0, and (z  2i) is a factor of the polynomial P(z).

220

20.6

POLYNOMIAL FUNCTIONS

[CHAP. 20

Prove that x  a is a factor of xn  an , if n is any positive integer. SOLUTION P(x) ¼ xn  an ; then P(a) ¼ an  an ¼ 0. Since P(a) ¼ 0, x  a is a factor of xn  an .

20.7

(a) Show that x5 þ a5 is exactly divisible by x þ a. (b) What is the remainder when y6 þ a6 is divided by y þ a? SOLUTION (a) P(x) ¼ x5 þ a5 ; then P(a) ¼ (a)5 þ a5 ¼ a5 þ a5 ¼ 0. Since P(a) ¼ 0, x5 þ a5 is exactly divisible by x þ a. (b) P(y) ¼ y6 þ a6 . Remainder ¼ P(a) ¼ (a)6 þ a6 ¼ a6 þ a6 ¼ 2a6 .

20.8

Show that x þ a is a factor of xn  an when n is an even positive integer, but not a factor when n is an odd positive integer. Assume a = 0. SOLUTION

P(x) ¼ xn  an : When n is even, P(a) ¼ (a)n  an ¼ an  an ¼ 0. Since P(a) ¼ 0, x þ a is a factor of xn  an when n is even. When n is odd, P(a) ¼ (a)n  an ¼ an  an ¼ 2an . Since P(a) = 0, xn  an is not exactly divisible by x þ a when n is odd (the remainder being 2an ).

20.9

Find the values of p for which (a) 2x3  px2 þ 6x  3p is exactly divisible by x þ 2, (b) (x4  p2 x þ 3  p) 4 (x  3) has a remainder of 4. SOLUTION (a) The remainder is 2(2)3  p(2)2 þ 6(2)  3p ¼ 16  4p  12  3p ¼ 28  7p ¼ 0. Then p ¼ 4. (b) The remainder is 34  p2 (3) þ 3  p ¼ 84  3p2  p ¼ 4. Then 3p2 þ p  80 ¼ 0, (p  5)(3p þ 16) ¼ 0 and p ¼ 5, 16=3.

20.10 By synthetic division, determine the quotient and remainder in the following. (3x5  4x4  5x3  8x þ 25) 4 (x  2) SOLUTION 2j

3  4  5 þ 0  8 þ 25 6 þ 4  2  4  24 3 þ 2  1  2  12 þ 1

Quotient: 3x4 þ 2x3  x2  2x  12 Remainder: 1

The top row of figures gives the coefficients of the dividend, with zero as the coefficient of the missing power of x (0x2 ). The 2 at the extreme left is the second term of the divisor with the sign changed (since the coefficient of x in the divisor is 1). The first coefficient of the top row, 3, is written first in the third row and then multiplied by the 2 of the divisor. The product 6 is placed first in the second row and added to the 24 above it to give 2, which is the next figure in the

CHAP. 20]

221

POLYNOMIAL FUNCTIONS

third row. This 2 is then multiplied by the 2 of the divisor. The product 4 is placed in the second row and added to the 25 above it to give the 21 in the third row; etc. The last figure in the third row is the Remainder, while all the figures to its left constitute the coefficients of the Quotient. Since the dividend is of the 5th degree and the divisor of the 1st degree, the quotient is of the 4th degree. The answer may be written: 3x4 þ 2x3  x2  2x  12 þ

1 : x2

20.11 (x4  2x3  24x2 þ 15x þ 50) 4 (x þ 4) SOLUTION 4 j

1  2  24 þ 15 þ 50  4 þ 24  0  60 1  6 þ 0 þ 15  10

Answer: x3  6x2 þ 15 

10 xþ4

20.12 (2x4  17x2  4) 4 (x þ 3) SOLUTION 3 j

2 þ 0  17 þ 0  4  6 þ 18  3 þ 9 26þ 13þ5

Answer: 2x3  6x2 þ x  3 þ

5 xþ3

20.13 (4x3  10x2 þ x  1) 4 (x  1=2) SOLUTION

 1=2  4  10 þ 1  1 þ 2  4  3=2 4  8  3  5=2

Answer: 4x2  8x  3 

5 2x  1

20.14 Given P(x) ¼ x3  6x2  2x þ 40, compute (a) P(5) and (b) P(4) using synthetic division. SOLUTION (a) 5 j

1  6  2 þ 40  5 þ 55  265

(b) 4 j 1  6  2 þ 40 þ 4  8  40

1  11 þ 53  225 P(5) ¼ 225

1  2  10 þ 0 P(4) ¼ 0

20.15 Given that one root of x3 þ 2x2  23x  60 ¼ 0 is 5, solve the equation. SOLUTION 5j

1 þ 2  23  60 þ 5 þ 35 þ 60

Divide x3 þ 2x2  23x  60 by x  5:

1 þ 7 þ 12 þ 0 The depressed equation is x2 þ 7x þ 12 ¼ 0, whose roots are 23, 24. The three roots are 5, 23, 24.

20.16 Two roots of x4  2x2  3x  2 ¼ 0 are 21 and 2. Solve the equation. SOLUTION 1 j

1þ0232 1þ1þ1þ2 1112þ0

Divide x4  2x2  3x  2 by x þ 1:

222

POLYNOMIAL FUNCTIONS

[CHAP. 20

The first depressed equation is x3  x2  x  2 ¼ 0. 2j 1112 þ2þ2þ2 Divide x3  x2  x  2 by x  2: 1þ1þ1þ0 1 1 pffiffiffi The second depressed equation is x2 þ x þ 1 ¼ 0, whose roots are  + i 3. 2 2 1 1 pffiffiffi The four roots are 1, 2,  + i 3. 2 2

20.17 Determine the roots of each of the following equations. (a) (b) (c) (d) (e) (f)

(x  1)2 (x þ 2)(x þ 4) ¼ 0. (2x þ 1)(3x  2)3 (2x  5) ¼ 0. x3 (x2  2x p ffiffiffi 15) ¼ 0. pffiffiffi (x þ 1 þ 3)(x þ 1  3)(x  6) ¼ 0. ½(x  i)(x þ i)3 (x þ 1)2 ¼ 0. 3(x þ m)4 (5x  n)2 ¼ 0:

Ans. 1 as a double root, 22, 24 21/2, 2/3 as a triple root, 5/2 0 as a p triple ffiffiffi root, 5,p23 ffiffiffi (1  3), (1 þ 3), 6 +i as triple roots, 21 as a double root m as a quadruple root, n/5 as a double root

20.18 Write the equation having only the following roots. (a) 5, 1, 23;

(b) 2, 1=4, 1=2;

(c) +2, 2 +

pffiffiffi 3;

(d ) 0, 1 + 5i.

SOLUTION (a) (x  5)(x  1)(x þ 3) ¼ 0 or x3  3x2  13x þ 15 ¼ 0:    1 1 5x2 11x 1  (b) (x  2) x þ xþ ¼ 0 or x3   ¼ 0 or 8x3  10x2  11x  2 ¼ 0; 4 4 2 8 4 which has integral coefficients. pffiffiffi pffiffiffi pffiffiffi pffiffiffi (c) (x  2)(x þ 2)½x  (2  3) ½x  (2 þ 3) ¼ (x2  4)½(x  2) þ 3 ½(x  2)  3 ¼ (x2  4)½(x  2)2  3 ¼ (x2  4)(x2  4x þ 1) ¼ 0,

or

x4  4x3  3x2 þ 16x  4 ¼ 0:

(d ) x½x  (1 þ 5i) ½x  (1  5i) ¼ x½(x  1)  5i ½(x  1) þ 5i ¼ x½(x  1)2 þ 25 ¼ x(x2  2x þ 26) ¼ 0,

or

x3  2x2 þ 26x ¼ 0:

20.19 Form the equation with integral coefficients having only the following roots. 1 1 (a) 1, ,  ; 2 3

3 2 (b) 0, , , 1; 4 3

(c) +3i; +

1pffiffiffi 2; 2

(d ) 2 as a triple root, 1:

SOLUTION (a) (x  1)(2x  1)(3x þ 1) ¼ 0 or 6x3  7x2 þ 1 ¼ 0 (b) x(4x  3)(3x  2)(x þ 1) ¼ 0 or 12x4  5x3  11x2 þ 6x ¼ 0      1pffiffiffi 1pffiffiffi 1 2 2 ¼ 0, 2 x 2 ¼ (x þ 9) x  (c) (x  3i)(x þ 3i) x  2 2 2 or

(x2 þ 9)(2x2  1) ¼ 0,

2x4 þ 17x2  9 ¼ 0

(d ) (x  2)3 (x þ 1) ¼ 0 or

x4  5x3 þ 6x2 þ 4x  8 ¼ 0

20.20 Each given number is a root of a polynomial equation with real coefficients. What other number is a pffiffiffi root? (a) 2i, (b) 3 þ 2i, (c) 3  i 2. SOLUTION

(a) 2i,

(b) 3  2i,

pffiffiffi (c) 3 þ i 2

CHAP. 20]

223

POLYNOMIAL FUNCTIONS

20.21 Each given number is a root of a polynomial equation with rational coefficients. What other number is a pffiffiffi pffiffiffi pffiffiffi root? (a)  7, (b) 4 þ 2 3, (c) 5  12 2. SOLUTION pffiffiffi pffiffiffi (a) 7, (b) 4  2 3,

pffiffiffi (c) 5 þ 12 2

20.22 Criticize the validity of each of the following conclusions. (a) x3 þ 7x  6i ¼ 0 has x ¼ i as a root; hence x ¼ i is a root. pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi (b) x3 þ (1  2 3)x2 þ (5  2 3)x þ 5 ¼ 0 has 3  i 2 as a root; hence 3 þ i 2 is a root. p ffiffi ffi p ffiffi ffi p ffiffi ffi pffiffiffi (c) x4 þ (1  2pffiffiffi2)x3 þ (4  2 2)x2 þ (3  4 2)x þ 1 ¼ 0 has x ¼ 1 þ 2 as a root; hence x ¼ 1  2 is a root. SOLUTION (a) x ¼ i is not necessarily a root, since not all the coefficients of the given equation are real. By substitution it is, in fact, found that x ¼ i is not a root. (b) The conclusion pffiffiffi is valid, since the given equation has real coefficients. (c) x ¼ 1  2 2 is not necessarily a root, pffiffiffisince not all the coefficients of the given equation are rational. By substitution it is found that x ¼ 1  2 is not a root.

20.23 Write the polynomial equation of lowest degree with real coefficients having 2 and 1 2 3i as two of its roots. SOLUTION (x  2)½x  (1  3i) ½x  (1 þ 3i) ¼ (x  2)(x2  2x þ 10) ¼ 0 or

x3  4x2 þ 14x  20 ¼ 0

20.24 Form the polynomial equation of lowest degree with rational coefficients having 1 þ two of its roots. SOLUTION ½x  (1 þ

pffiffiffi pffiffiffi 5) ½x  (1  5)(x þ 6) ¼ (x2 þ 2x  4)(x þ 6) ¼ 0 or

pffiffiffi 5 and 26 as

x3 þ 8x2 þ 8x  24 ¼ 0

20.25 Form the quartic polynomial equation with rational coefficients having as two of its roots pffiffiffi pffiffiffi (a) 5i and 6, (b) 2 þ i and 1  3. SOLUTION

pffiffiffi pffiffiffi 6)(x þ 6) ¼ (x2 þ 25)(x2  6) ¼ 0 or x4 þ 19x2  150 ¼ 0 pffiffiffi pffiffiffi (b) ½x  (2 þ i) ½x  (2  i) ½x  (1  3) ½x  (1 þ 3) ¼ (x2  4x þ 5)(x2  2x  2) ¼ 0 (a) (x þ 5i)(x  5i)(x 

or

x4  6x3 þ 11x2  2x  10 ¼ 0

20.26 Find the four roots of x4 þ 2x2 þ 1 ¼ 0. SOLUTION x4 þ 2x2 þ 1 ¼ (x2 þ 1)2 ¼ ½(x þ i)(x  i)2 ¼ 0:

The roots are i, i, i, i:

20.27 Solve x4  3x3 þ 5x2  27x  36 ¼ 0, given that one root is a pure imaginary number of the form bi where b is real.

224

POLYNOMIAL FUNCTIONS

[CHAP. 20

SOLUTION Substituting bi for x, b4 þ 3b3 i  5b2  27bi  36 ¼ 0. Equating real and imaginary parts to zero: b4  5b2  36 ¼ 0, (b2  9)(b2 þ 4) ¼ 0 and b ¼ +3 since b is real; 3b3  27b ¼ 0, 3b(b2  9) ¼ 0 and b ¼ 0, +3: The common solution is b ¼ +3; hence two roots are +3i and (x  3i)(x þ 3i) ¼ x2 þ 9 is a factor of x  3x3 þ 5x2  27x  36. By division the other factor is x2  3x  4 ¼ (x  4)(x þ 1), and the other two roots are 4, 21. The four roots are +3i, 4; 1. 4

20.28 Formpffiffithe polynomial ffiffiffi pffiffiequation ffi pffiffiffiffiffiffiffi of lowest degree with rational coefficients, one of whose roots is ffi p (a) 3  2, (b) 2 þ 1. SOLUTION pffiffiffi pffiffiffi (a) Let x ¼ 3  2.

pffiffiffi pffiffiffi pffiffiffi Squaring both sides, x2 ¼ 3  2 6 þ 2 ¼ 5  2 6 and x2  5 ¼ 2 6:

Squaring again, x4  10x2 þ 25 ¼ 24 and x4  10x2 þ 1 ¼ 0: pffiffiffi pffiffiffiffiffiffiffi (b) Let x ¼ 2 þ 1. pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi Squaring both sides, x2 ¼ 2 þ 2 2  1 ¼ 1 þ 2 2 and x2  1 ¼ 2 2: Squaring again, x4  2x2 þ 1 ¼ 8 and x4  2x2 þ 9 ¼ 0:

20.29 (a) Write the polynomial equation of lowest degree with constant (real or complex) coefficients having the roots 2 and 1  3i. Compare with Problem 20.23 above. (b) Write p the ffiffiffi polynomial equation of lowest degree with real coefficients having the roots 26 and 1 þ 5. Compare with Problem 20.24 above. SOLUTION (a) (x  2)½x  (1  3i) ¼ 0 or x2  3(1  i)x þ 2  6i ¼ 0 pffiffiffi pffiffiffi pffiffiffi (b) (x þ 6)½x  (1 þ 5) ¼ 0 or x2 þ (7  5)x  6( 5  1) ¼ 0

20.30 Obtain the rational roots, if any, of each of the following polynomial equations. (a) x4  2x2  3x  2 ¼ 0 The rational roots are limited to the integral factors of 2, which are +1, +2. Testing these values for x in order, þ1, 1, þ2, 2, by synthetic division or by substitution, we find that the only rational roots are 21 and 2. (b) x3  x  6 ¼ 0 The rational roots are limited to the integral factors of 6, which are +1, +2, +3, +6. Testing these values for x in order, þ1, 1, þ2, 2, þ3, 3, þ6, 6, the only rational root obtained is 2. (c) 2x3 þ x2  7x  6 ¼ 0 If b/c (in lowest terms) is a rational root, the only possible values of b are +1, +2, +3, +6; and the only possible values of c are +1, +2. Hence the possible rational roots are limited to the following numbers: +1, +2, +3, +6, +1=2, +3=2. Testing these values for x, we obtain 1, 2, 3=2 as the rational roots.

CHAP. 20]

225

POLYNOMIAL FUNCTIONS

(d ) 2x4 þ x2 þ 2x  4 ¼ 0 If b/c is a rational root, the values of b are limited to +1, +2, +4; and the values of c are limited to +1, +2. Hence the possible rational roots are limited to the numbers +1, +2, +4, +1=2. Testing these values for x, we find that there are no rational roots. 20.31 Solve the polynomial equation x3  2x2  31x þ 20 ¼ 0. SOLUTION Any rational root of this equation is an integral factor of 20. Then the possibilities for rational roots are: +1, +2, +4, +5, +10, +20. Testing these values for x by synthetic division, we find that 5 is a root. 5 j 1  2  31 þ 20  5 þ 35  20 17þ 4þ 0 The depressed equation x2  7x þ 4 ¼ 0 has irrational roots 7=2 + pffiffiffiffiffi Hence the three roots of the given equation are 5, 7=2 + 33=2.

pffiffiffiffiffi 33=2.

20.32 Solve the polynomial equation 2x4  3x3  7x2  8x þ 6 ¼ 0. SOLUTION If b/c is a rational root, the only possible values of b are +1, +2, +3, +6; and the only possible values of c are +1, +2. Hence the possibilities for rational roots are +1, +2, +3, +6, +1=2, +3=2. Testing these values of x by synthetic division, we find that 3 is a root. 3j 2378þ6 þ6þ9þ66 2þ3þ22þ0 The first depressed 2x3 þ 3x2 þ 2x  2 ¼ 0 is tested and 1/2 is obtained as a root.  1=2  2 þ 3 þ 2  2 þ1þ2þ2 2þ4þ4þ0 The second depressed equation 2x2 þ 4x þ 4 ¼ 0 or x2 þ 2x þ 2 ¼ 0 has the nonreal roots 1 + i: The four roots are 3, 1=2, 1+i.

20.33 Prove that

pffiffiffi pffiffiffi 3 þ 2 is an irrational number.

SOLUTION pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi Let x ¼ 3 þ 2; then x2 ¼ ( 3 þ 2)2 ¼ 3 þ 2 6 þ 2 ¼ 5 þ 2 6 and x2  5 ¼ 2 6. Squaring again, x4  10x2 þ 25 ¼ 24 or x4  10x2 þ 1 ¼ 0. The only possible pffiffiffi roots of this equation pffiffiffi rational are +1. Testing these values, we find that there is no rational root. Hence x ¼ 3 þ 2 is irrational.

226

POLYNOMIAL FUNCTIONS

[CHAP. 20

20.34 Graph P(x) ¼ x3 þ x  3. From the graph determine the number of positive, negative and nonreal roots of x3 þ x  3 ¼ 0. SOLUTION x

23

22

21

0

1

2

3

4

P(x)

233

213

25

23

21

7

27

65

From the graph it is seen that there is one positive and no negative real root (see Fig. 20-1). Hence there are two conjugate nonreal roots.

20.35 Find upper and lower limits of the real roots of (a) x3  3x2 þ 5x þ 4 ¼ 0, (b) x3 þ x2  6 ¼ 0:

Fig. 20-1 SOLUTION (a) The possible rational roots are +1, +2, +4. Testing for upper limit. 1j

13þ5þ4

2j

13þ5þ 4

þ12þ3

þ22þ 6

3j 13þ5þ 4 þ 3 þ 0 þ 15

12þ3þ7

1  1 þ 3 þ 10

1 þ 0 þ 5 þ 19

Since all the numbers in the third row of the synthetic division of P(x) by x  3 are positive (or zero), an upper limit of the roots is 3, i.e., no root is greater than 3. Testing for lower limit. 1 j 1  3 þ 5 þ 4 1þ49 14þ95 Since the numbers in the third row are alternately positive and negative, 1 is a lower limit of the roots, i.e., no root is less than 1. (b) The possible rational roots are +1, +2, +3, +6.

CHAP. 20]

POLYNOMIAL FUNCTIONS

227

Testing for upper limit. 1j

1þ1þ06 þ1þ2þ2

2j 1þ1þ0 6 þ 2 þ 6 þ 12

1þ2þ24

1þ3þ6þ 6

Hence 2 is an upper limit of the roots. Testing for lower limit. 1 j

1þ1þ06 10þ0 1þ0þ06

Since all the numbers of the third row are alternately positive and negative (or zero), a lower limit to the roots is 1.

20.36 Determine the rational roots of 4x3 þ 15x  36 ¼ 0 and thus solve the equation completely.

SOLUTION The possible rational roots are +1, +2, +3, +4, +6, +9, +12, +18, +36, +1=2, +3=2, +9=2, +1=4, +3=4, +9=4. To avoid testing all of these possibilities, find upper and lower limits of the roots. Testing for upper limit. 1j

2j

4 þ 0 þ 15  36 þ 4 þ 4 þ 19 4 þ 4 þ 19  17

4 þ 0 þ 15  36 þ 8 þ 16 þ 62 4 þ 8 þ 31 þ 26

Hence no (real) root is greater than or equal to 2. Testing for lower limit. 1 j 4 þ 0 þ 15  36  4 þ 4  19 4  4 þ 19  55 Hence no real root is less than or equal to 1. The only possible rational roots greater than 1 and less than 2 are þ1, +1=2, +3=2, +1=4, +3=4. Testing these we find that 3/2 is the only rational root.  3=2 

4 þ 0 þ 15  36 þ 6 þ 9 þ 36 4 þ 6 þ 24 þ 0

pffiffiffiffiffi 87 3 i: The other roots are solutions of 4x þ 6x þ 24 ¼ 0 or 2x þ 3x þ 12 ¼ 0, i.e., x ¼  + 4 4 2

2

228

POLYNOMIAL FUNCTIONS

[CHAP. 20

20.37 Employing Descartes’ Rule of Signs, what may be inferred as to the number of positive, negative and nonzero roots of the following equations? (a) 2x3 þ 3x2  13x þ 6 ¼ 0 (b) x4  2x2  3x  2 ¼ 0 (c) x2  2x þ 7 ¼ 0

(d ) 2x4 þ 7x2 þ 6 ¼ 0 (e) x4  3x2  4 ¼ 0 ( f ) x3 þ 3x  14 ¼ 0

(g) x6 þ x3  1 ¼ 0 (h) x6  3x2  4x þ 1 ¼ 0

SOLUTION (a) There are 2 variations of sign in P(x) ¼ 2x3 þ 3x2  13x þ 6. There is 1 variation of sign in P(x) ¼ 2x3 þ 3x2 þ 13x þ 6. Hence there are at most 2 positive roots and 1 negative root. The roots may be: (1) 2 positive, 1 negative, 0 nonreal; or (2) 0 positive, 1 negative, 2 nonreal. (Nonreal roots occur in conjugate pairs.) (b) There is 1 variation of sign in P(x) ¼ x4  2x2  3x  2, and 3 variations P(x) ¼ x4  2x2 þ 3x  2. Hence there are at most 1 positive root and 3 negative roots. The roots may be: (1) 1 positive, 3 negative, 0 nonreal; or (2) 1 positive, 1 negative, 2 nonreal.

of

sign

in

(c) There are 2 variations of sign in P(x) ¼ x2  2x þ 7, and no variation of sign in P(x) ¼ x2 þ 2x þ 7: Hence the roots may be: (1) 2 positive, 0 negative, 0 nonreal; or (2) 0 positive, 0 negative, 2 nonreal. (d ) Neither P(x) ¼ 2x4 þ 7x2 þ 6 nor P(x) ¼ 2x4 þ 7x2 þ 6 has a variation of sign. Hence all 4 roots are nonreal, since P(0)= 0. (e) There is 1 variation of sign in P(x) ¼ x4  3x2  4 ¼ 0, and 1 variation of sign in P(x) ¼ x4  3x2  4. Hence the roots are: 1 positive, 1 negative, 2 nonreal. ( f ) There is 1 variation of sign in P(x) ¼ x3 þ 3x  14, and no variation in P(x) ¼ x3  3x  14. Hence the roots are: 1 positive, 2 nonreal. (g) There is 1 variation of sign in P(x) ¼ x6 þ x3  1, and 1 variation in P(x) ¼ x6  x3  1: Hence the roots are: 1 positive, 1 negative, 4 nonreal. (h) There are 2 variations of sign in P(x) ¼ x6  3x2  4x þ 1, and 2 variations of sign in P(x) ¼ x6  3x2 þ 4x þ 1. Hence the roots may be: (1) 2 positive, 2 negative, 2 nonreal; (3) 0 positive, 2 negative, 4 nonreal; (2) 2 positive, 0 negative, 4 nonreal; (4) 0 positive, 0 negative, 6 nonreal.

20.38 Determine the nature of the roots of xn  1 ¼ 0 when n is a positive integer and (a) n is even, (b) n is odd. SOLUTION (a) P(x) ¼ xn  1 has 1 variation of sign, and P(x) ¼ xn  1 has 1 variation of sign. Hence the roots are: 1 positive, 1 negative, (n  2) nonreal. (b) P(x) ¼ xn  1 has 1 variation of sign, and P(x) ¼ xn  1 has no variation of sign. Hence the roots are: 1 positive, 0 negative, (n  1) nonreal.

20.39 Obtain the rational roots, if any, of each equation, making use of Descartes’ Rule of Signs. (a) x3  x2 þ 3x  27 ¼ 0, (b) x3 þ 2x þ 12 ¼ 0,

(c) 2x5 þ x  66 ¼ 0, (d ) 3x4 þ 7x2 þ 6 ¼ 0.

SOLUTION (a) By Descartes’ Rule of Signs, the equation has 3 or 1 positive roots and no negative root. Hence the rational roots are limited to positive integral factors of 27, which are 1, 3, 9, 27. Testing these values for x, the only rational root obtained is 3. (b) By the rule of signs, the equation has no positive root and 1 negative root. Hence the rational roots are limited to the negative integral factors of 12, i.e., to 1, 2, 3, 4, 6, 12. Testing these values for x, we obtain 2 as the only rational root.

CHAP. 20]

229

POLYNOMIAL FUNCTIONS

(c) By the rule of signs, the equation has 1 positive root and no negative root. Hence the rational roots are limited to positive rational numbers of the form b/c where b is limited to integral factors of 66 and c is limited to integral factors of 2. The possible rational roots are then 1, 2, 3, 6, 11, 22, 33, 66, 1/2, 3/2, 11/2, 33/2. Testing these values for x, we obtain 2 as the only rational root. (d ) The equation has no real root, since neither P(x) ¼ 3x4 þ 7x2 þ 6 nor P(x) ¼ 3x4 þ 7x2 þ 6 has a variation of sign and P(0)=0. Hence all of its four roots are nonreal.

20.40 Use the Intermediate Value Theorem to isolate each of the real zeros of P(x) between two consecutive integers. (a) P(x) ¼ 3x3  8x2  8x þ 8 (b) P(x) ¼ 5x3  4x2  10x þ 8 SOLUTION (a) For P(x) ¼ 3x3  8x2  8x þ 8, we find the upper and lower limits of the real zeros. 0j

1j

388þ8 þ0þ0þ0 388þ8

3j

38 8þ 8 þ 3  5  13

2j

3  2  12  16

3  5  13  5

388þ 8 þ 9 þ 3  15

4j 3 8 8þ 8 þ 12 þ 16 þ 32

3þ15 7

3 þ 4 þ 8 þ 40

38 8þ 8 þ 6  4  24

Since the quotient row when synthetic division is done with 4 is all positive, the upper limit of the real zeros of P(x) is 4. 1 j 3  8  8 þ 8  3 þ 11  3

2 j 3  8  8 þ 8  6 þ 28  40

3  11 þ 3 þ 5

3  14 þ 14  32

Since the quotient row when synthetic division is done with 2 alternates in sign, the lower limit of the real zeros of P(x) is 2. We now examine the interval from 2 to 4 to isolate the real zeros of P(x) between consecutive integers. Since P(0) ¼ 8 and P(1) ¼ 5 there is a real zero between 0 and 1. Since P(3) ¼ 7 and P(4) ¼ 40 there is a real zero between 3 and 4. Since P(1) ¼ 5 and P(2) ¼ 32, there is a real zero between 2 and 1. The real zeros of P(x) ¼ 3x3  8x2  8x þ 8 are between 2 and 1, 0 and 1, and 3 and 4. (b) For P(x) ¼ 5x3  4x2  10x þ 8, we find the upper and lower limits of the real zeros. 0j

5  4  10 þ 8 þ0þ 0þ0

1j

5  4  10 þ 8

5  4  10 þ 8 þ5þ 19 5þ1 91

2j

5  4  10 þ 8 þ 10 þ 12 þ 4 5 þ 6 þ 2 þ 12

The upper limit for the zeros of P(x) is 2. 1 j

5  4  10 þ 8 5þ 9þ1 59 1þ9

The lower limit of the real zeros of P(x) is 2.

2 j

5  4  10 þ 8  10 þ 28  36 5  14 þ 18  28

230

POLYNOMIAL FUNCTIONS

[CHAP. 20

We now examine the interval from 2 to 2 to isolate the real zeros of P(x). Since P(0) ¼ 8 and P(1) ¼ 1, there is a real zero between 0 and 1. Since P(1) ¼ 1 and P(2) ¼ 12, there is a real zero between 1 and 2. Since P(1) ¼ 9 and P(2) ¼ 28, there is a real zero between 2 and 1. Thus, the real zeros of P(x) ¼ 5x3  4x2  10x þ 8 are located between 2 and 1, 0 and 1, and 1 and 2.

20.41 Approximate a real zero of P(x) ¼ x3  x  5 to two decimal places. SOLUTION By Descartes’ Rule of Signs, P(x) ¼ x3  x  5 has 1 positive real zero and 2 or 0 negative real zeros. Now we determine the upper limit of real zeros of P(x). 0j 1þ015 þ0þ0þ0

1j 1þ015 þ1þ1þ0

1þ015

1þ1þ05

2j

1þ015 þ2þ4þ6 1þ2þ3þ1

The upper limit on the real zeros of P(x) is 2. Since P(1) ¼ 25 and P(2) ¼ 1, the positive real zero is between 1 and 2. We now determine the tenths interval of the zero. Using synthetic division, we determine the tenths values until we find two values with different signs. Since P(2) ¼ 1 is closer to 0 than P(1) ¼ 5, we start with x ¼ 1.9. Since P(1.9) ¼ 20.041 and P(2.0) ¼ 1, the real zero is between 1.9 and 2.0. Since P(1.90) ¼ 20.41 is closer to 0 than P(2.0) ¼ 1, we look for the hundredths interval starting with x ¼ 1.91. P(1.91) ¼ 0.579. Since P(1.90) ¼ 20.041 and P(1.91) ¼ 0.058, there is a real zero between 1.90 and 1.91. We now determine P(1.905) to decide whether the zero rounds off to 1.90 or 1.91. P(1.905) ¼ 0.008. Since P(1.900) is negative and P(1.905) is positive, the zero is between 1.900 and 1.905. When rounded to two decimal places, all numbers in this interval round to 1.90. Thus, rounded to two decimal places, a real zero of P(x) ¼ x3  x  5 is 1.90.

20.42 Approximate

ffiffiffi p 3 3 to three decimal places.

SOLUTION pffiffiffi Let x ¼ 3 3, so x3 ¼ 3 and P(x) ¼ x3  3 ¼ 0. By Descartes’ Rule of Signs, P(x) has one positive real zero and no negative real zeros. Since P(1) ¼ 2 and P(2) ¼ 5, the zero is between 1 and 2. Since P(1) is closer to 0 than P(2), locate the tenths interval by evaluating P(x) from x ¼ 1 to x ¼ 2 starting with x ¼ 1.1. Once a sign change in P(x) is found we stop. x

1.0

1.1

1.2

1.3

1.4

1.5

P(x)

22

21.669

21.272

20.803

20.256

0.375

Since P(1.4) is negative and P(1.5) is positive, the real zero is between 1.4 and 1.5. Now determine the hundredths interval by exploring the values of P(x) in the interval from x ¼ 1.40 to x ¼ 1.50. x

1.40

1.41

1.42

1.43

1.44

1.45

P(x)

20.256

20.197

20.137

20.076

20.014

0.049

Since P(1.4) is negative and P(1.5) is positive, the zero is between 1.4 and 1.5. The next step is to determine the thousandths interval for the zero by exploring the values of P(x) in the interval between x ¼ 1.440 and x ¼ 1.450. x P(x)

1.440

1.441

1.442

1.443

20.014

20.008

20.002

0.005

CHAP. 20]

231

POLYNOMIAL FUNCTIONS

Since P(1.442) is negative and P(1.443) is positive, the zero is between 1.442 and 1.443. Since P(1.4425) ¼ 0.002, the real zero is between 1.4420 and 1.4425. All values in the interval from 1.4420 to 1.4425 round to 1.442 to three decimal places. pffiffiffi Therefore, 3 3 ¼ 1:442 to three decimal places.

Supplementary Problems

20.43

pffiffiffi If P(x) ¼ 2x  x  x þ 2, find (a) P(0), (b) P(2), (c) P(1), (d) P( 12 ), (e) P( 2).

20.44

Determine the remainder in each of the following.

3

2

(a) (2x5  7) 4 (x þ 1) (b) (x3 þ 3x2  4x þ 2) 4 (x  2) (c) (3x3 þ 4x  4) 4 (x  12 )

(d ) (4y3 þ y þ 27) 4 (2y þ 3) pffiffiffiffiffiffiffi (e) (x12 þ x6 þ 1) 4 (x  1) ( f ) (2x33 þ 35) 4 (x þ 1)

20.45

Prove that x þ 3 is a factor of x3 þ 7x2 þ 10x  6 and that x ¼ 3 is a root of the equation x3 þ 7x2 þ 10x  6 ¼ 0.

20.46

Determine which of the following numbers are roots of the equation y4 þ 3y3 þ 12y  16 ¼ 0: (a) 2,

20.47

(b) 4,

(c) 3,

(d ) 1,

(e) 2i.

Find the values of k for which (a) 4x3 þ 3x2  kx þ 6k is exactly divisible by x þ 3, (b) x5 þ 4kx  4k2 ¼ 0 has the root x ¼ 2.

20.48

By synthetic division determine the quotient and remainder in each of the following. (c) (y6  3y5 þ 4y  5) 4 (y þ 2) (a) (2x3 þ 3x2  4x  2) 4 (x þ 1) (b) (3x5 þ x3  4) 4 (x  2)

(d ) (4x3 þ 6x2  2x þ 3) 4 (2x þ 1)

20.49

If P(x) ¼ 2x4  3x3 þ 4x  4, compute P(2) and P(3) using synthetic division.

20.50

Given that one root of x3  7x  6 ¼ 0 is 1, find the other two roots.

20.51

Show that 2x4  x3  3x2  31x  15 ¼ 0 has roots 3,  12. Find the other roots.

20.52

Find the roots of each equation. (a) (x þ 3)2 (x  2)3 (x þ 1) ¼ 0 (b) 4x4 (x þ 2)4 (x  1) ¼ 0

(c) (x2 þ 3x þ 2)(x2  4x þ 5) ¼ 0 (d) (y2 þ 4)2 (y þ 1)2 ¼ 0

20.53

Form equations with integral coefficients having only the following roots. (b) 0, 4, 2=3, 1 (c) +3i, double root 2 (d ) 1 + 2i, 2 + i (a) 2, 3,  12

20.54

Form an equation whose only roots are 1 +

20.55

Write the equation of lowest possible degree with integral coefficients having the given roots. pffiffiffi pffiffiffi pffiffiffi (a) 1, 0, i (b) 2 þ i (c) 1 + 3, 1=3 (d ) 2, i 3 (e) 2, i ( f ) i=2, 6=5

20.56

In the equation x3 þ ax2 þ bx þ a ¼ 0, a and b are real numbers. If x ¼ 2 þ i is a root of the equation, find a and b.

20.57

Write an equation of lowest degree with integral coefficients having

pffiffiffi 2  1 as a double root.

20.58

Write an equation of lowest degree with integral coefficients having

pffiffiffi 3 þ 2i as a root.

pffiffiffi pffiffiffi 2, 1 + i 3.

232

20.59

POLYNOMIAL FUNCTIONS

Solve each equation, given the indicated root. pffiffiffi (a) x4 þ x3  12x2 þ 32x  40 ¼ 0; 1  i 3 pffiffiffi (b) 6x4  11x3 þ x2 þ 33x  45 ¼ 0; 1 þ i 2

(c) x3  5x2 þ 6 ¼ 0; 3 

[CHAP. 20

pffiffiffi 3

(d ) x4  4x3 þ 6x2  16x þ 8 ¼ 0; 2i

20.60

Obtain the rational roots, if any, of each equation. (c) 2x4  x3 þ 2x2  2x  4 ¼ 0 (a) x4 þ 2x3  4x2  5x  6 ¼ 0 3 (b) 4x  3x þ 1 ¼ 0 (d ) 3x3 þ x2  12x  4 ¼ 0

20.61

Solve each equation. (a) x3  x2  9x þ 9 ¼ 0 (b) 2x3  3x2  11x þ 6 ¼ 0 (c) 3x3 þ 2x2 þ 2x  1 ¼ 0

(d) 4x4 þ 8x3  5x2  2x þ 1 ¼ 0 (e) 5x4 þ 3x3 þ 8x2 þ 6x  4 ¼ 0 ( f ) 3x5 þ 2x4  15x3  10x2 þ 12x þ 8 ¼ 0

pffiffiffi pffiffiffi pffiffiffi 5  2 and (b) 5 2 are irrational numbers.

20.62

Prove that (a)

20.63

If P(x) ¼ 2x3  3x2 þ 12x  16, determine the number of positive, negative and nonreal roots.

20.64

Locate between two successive integers the real roots of x4  3x2  6x  2 ¼ 0: Find the least positive root of the equation accurate to two decimal places.

20.65

Find upper and lower limits for the real roots of each equation. (b) 2x4 þ 5x2  6x  14 ¼ 0 (a) x3  3x2 þ 2x  4 ¼ 0

20.66

Find the rational roots of 2x3  5x2 þ 4x þ 24 ¼ 0 and thus solve the equation completely.

20.67

Using Descartes’ Rule of Signs, what may be inferred as to the number of positive, negative and nonreal roots of the following equations? (c) x5 þ 4x3  3x2  x þ 12 ¼ 0 (a) 2x3 þ 3x2 þ 7 ¼ 0 3 2 (b) 3x  x þ 2x  1 ¼ 0 (d ) x5  3x  2 ¼ 0

20.68

Given the equation 3x4  x3 þ x2  5 ¼ 0, determine (a) the maximum number of positive roots, (b) the minimum number of positive roots, (c) the exact number of negative roots, and (d) the maximum number of nonreal roots.

20.69

Given the equation 5x3 þ 2x  4 ¼ 0, how many roots are (a) negative, (b) real?

20.70

Does the equation x6 þ 4x4 þ 3x2 þ 16 ¼ 0 have (a) 4 nonreal and 2 real roots, (b) 4 real and 2 nonreal roots, (c) 6 nonreal roots, or (d) 6 real roots?

20.71

(a) How many positive roots has the equation x6  7x2  11 ¼ 0? (b) How many complex roots has the equation x7 þ x4  x2  3 ¼ 0? (c) Show that x6 þ 2x3 þ 3x  4 ¼ 0 has exactly 4 nonreal roots. (d ) Show that x4 þ x3  x2  1 ¼ 0 has only one negative root.

20.72

Solve completely each equation. (a) 8x3  20x2 þ 14x  3 ¼ 0

(c) 4x3 þ 5x2 þ 2x  6 ¼ 0

(b) 8x  14x  9x þ 11x  2 ¼ 0

(d ) 2x4  x3  23x2 þ 18x þ 18 ¼ 0

4

20.73

3

2

Approximate the indicated root of each equation to the specified accuracy. (a) 2x3 þ 3x2  9x  7 ¼ 0; positive root, to the nearest tenth (b) x3 þ 9x2 þ 27x  50 ¼ 0; positive root, to the nearest hundredth (c) x3  3x2  3x þ 18 ¼ 0; negative root, to the nearest tenth

CHAP. 20]

233

POLYNOMIAL FUNCTIONS

(d ) x3 þ 6x2 þ 9x þ 17 ¼ 0; negative root, to the nearest tenth (e) x5 þ x4  27x3  83x2 þ 50x þ 162 ¼ 0; root between 5 and 6, to the nearest hundredth ( f ) x4  3x3 þ x2  7x þ 12 ¼ 0; root between 1 and 2, to the nearest hundredth 20.74

In finding the maximum deflection of a beam of given length loaded in a certain way, it is necessary to solve the equation 4x3  150x2 þ 1500x  2871 ¼ 0. Find correct to the nearest tenth the root of the equation lying between 2 and 3.

20.75

The length of a rectangular box is twice its width, and its depth is one foot greater than its width. If its volume is 64 cubic feet, find its width to the nearest tenth of a foot.

20.76

Find

p ffiffiffiffiffi 3 20 correct to the nearest hundredth.

ANSWERS TO SUPPLEMENTARY PROBLEMS 20.43

(a) 2

(b) 12

(c) 0

20.44

(a) 9

20.46

4, 1 and 2i are roots

20.47

(a) k ¼ 9

20.48

(a) 2x2 þ x  5 þ

(c) 13=8

(b) 14

pffiffiffi (e) 3 2

(d ) 3/2

(d ) 12

(e) 1

( f ) 33

(b) k ¼ 4, 2 3 xþ1

(c) y5  5y4 þ 10y3  20y2 þ 40y  76 þ

(b) 3x4 þ 6x3 þ 13x2 þ 26x þ 52 þ

100 x2

20.49

12, 227

20.50

3, 2

20.51

1 + 2i

20.52

(a) double root 3, triple root 2, 1 (b) quadruple root 0, quadruple root 2, 1

20.53

(a) 2x3 þ 3x2  11x  6 ¼ 0 (b) 3x4 þ 7x3  18x2 þ 8x ¼ 0

20.54

x4  x2  10x  4 ¼ 0

20.55

(a) x4  x3 þ x2  x ¼ 0 (b) x2  4x þ 5 ¼ 0 (c) 3x3 þ 5x2  8x þ 2 ¼ 0

20.56

a ¼ 5, b ¼ 9

20.57

x4 þ 4x3 þ 2x2  4x þ 1 ¼ 0

20.58

x4 þ 2x2 þ 49 ¼ 0

(d ) 2x2 þ 2x  2 þ

5 2x þ 1

(c) 1, 2, 2 + i (d ) double roots +2i, double root 1

(c) x4  4x3 þ 13x2  36x þ 36 ¼ 0 (d ) x4  2x3 þ 2x2  10x þ 25 ¼ 0

(d) x3 þ 2x2 þ 3x þ 6 ¼ 0 (e) x4  x2  2 ¼ 0 ( f ) 20x3  24x2 þ 5x  6 ¼ 0

147 yþ2

234

POLYNOMIAL FUNCTIONS

20.59

pffiffiffi (a) 1 + i 3, 5, 2

20.60

(a) 3, 2

20.61

(a) 1, +3

pffiffiffi (b) 1 + i 2, 5=3, 3=2

(b) 1=2, 1=2, 1

(b) 3,  2; 1=2 (c) 1=3,  12 þ 12

(c) 3 +

(c) no rational root

[CHAP. 20

pffiffiffi 3, 1

(d ) +2i, 2 +

pffiffiffi 2

(d ) 1=3, +2

pffiffiffi (d ) + 12 , 1 + 2 pffiffiffi (e) 1, 2=5, + 2i

pffiffiffi 3i

( f ) +1, +2, 2=3

20.63

1 positive root, 0 negative roots, 2 nonreal roots.

20.64

Positive root between 2 and 3; negative root between 1 and 0; positive root ¼ 2.41 approx.

20.65

(a) Upper limit 3, lower limit 1

20.66

3=2, 2 + 2i

20.67

(a) 1 negative, 2 nonreal

(b) Upper limit 2, lower limit 2

(b) 3 positive or 1 positive, 2 nonreal (c) 1 negative, 2 positive, 2 nonreal or 1 negative, 4 nonreal (d ) 1 positive, 2 negative, 2 nonreal or 1 positive, 4 nonreal 20.68

(a) 3

(b) 1

20.69

(a) none

20.70

(c)

20.71

(a) one

20.72

(a) 1/2, 1/2, 3/2

20.73

(a) 1.9

20.74

2.5

20.75

2.9 ft

20.76

2.71

(c) 1

(d ) 2

(b) one

(b) four or six

(b) 1.25

(b) 2, 1, 1/4, 1/2 (c) 2:2

(c) 3/4, 1 + i

(d) 4:9

(e) 5.77

(d ) 3, 3/2, 2 + ( f ) 1.38

pffiffiffi 2

CHAPTER 21

Rational Functions

21.1

RATIONAL FUNCTIONS

A rational function is the ratio of two polynomial functions. If P(x) and Q(x) are polynomials, then a function of the form R(x) ¼ P(x)/Q(x) is a rational function where Q(x)=0. The domain of R(x) is the intersection of the domains of P(x) and Q(x). 21.2

VERTICAL ASYMPTOTES

If R(x) ¼ P(x)/Q(x), then values of x that make Q(x) ¼ 0 result in vertical asymptotes if P(x)=0. However, if for some value x ¼ a, P(a) ¼ 0 and Q(a) ¼ 0, then P(x) and Q(x) have a common factor of x  a. If R(x) is then reduced to lowest terms, the graph of R(x) has a hole in it where x ¼ a. A vertical asymptote for R(x) is a vertical line x ¼ k, k is a constant, that the graph of R(x) approaches but does not touch. R(k) is not defined because Q(k) ¼ 0 and P(k) = 0. The domain of R(x) is separated in distinct intervals by the vertical asymptotes of R(x). EXAMPLE 21.1.

What are the vertical asymptotes of R(x) ¼

2x  3 ? x2  4

R(x) ¼

2x  3 x2  4

Since

is undefined when x2  4 ¼ 0, x ¼ 2 and x ¼ 2 could result in vertical asymptotes. When x ¼ 2, 2x  3 = 0 and when x ¼ 2, 2x  3 = 0. Thus, the graph of R(x) has vertical asymptotes of x ¼ 2 and x ¼ 2:

21.3

HORIZONTAL ASYMPTOTES

A rational function R(x) ¼ P(x)/Q(x) has a horizontal asymptote y ¼ a if, as jxj increases without limit, R(x) approaches a. R(x) has at most one horizontal asymptote. The horizontal asymptote of R(x) may be found from a comparison of the degree of P(x) and the degree of Q(x). 235

236

RATIONAL FUNCTIONS

[CHAP. 21

(1) If the degree of P(x) is less than the degree of Q(x), then R(x) has a horizontal asymptote of y ¼ 0. (2) If the degree of P(x) is equal to the degree of Q(x), then R(x) has a horizontal asymptote of y ¼ an =bn , where an is the lead coefficient (coefficient of the highest degree term) of P(x) and bn is the lead coefficient of Q(x). (3) If the degree of P(x) is greater than the degree of Q(x), then R(x) does not have a horizontal asymptote. The graph of R(x) may cross a horizontal asymptote in the interior of its domain. This is possible since we are only concerned with how R(x) behaves as jxj increases without limit in determining the horizontal asymptote. EXAMPLE 21.2. (a) R(x) ¼

3x3 x2  1

What are the horizontal asymptotes of each rational function R(x)? (b) R(x) ¼

x x2  4

(c) R(x) ¼ R(x) ¼

(a) For

2x þ 1 3 þ 5x 3x3 , x2  1

the degree of the numerator 3x3 is 3 and the degree of the denominator is 2. Since the numerator exceeds the degree of the denominator, R(x) does not have a horizontal asymptote. (b) The degree of the numerator of R(x) ¼

x x2  1

is 1 and the degree of the denominator is 2, so R(x) has a horizontal asymptote of y ¼ 0. (c) The numerator and denominator of R(x) ¼

2x þ 1 3 þ 5x

each have degree 1. Since the lead coefficient of the numerator is 2 and the lead coefficient of the denominator is 5, R(x) has a horizontal asymptote of y ¼ 25.

21.4

GRAPHING RATIONAL FUNCTIONS

To graph a rational function R(x) ¼ P(x)/Q(x), we first determine the holes: values of x for which both P(x) and Q(x) are zero. After any holes are located, we reduce R(x) to lowest terms. The value of the reduced form of R(x) for an x that yields a hole is the y coordinate of the point that corresponds to the hole. Once R(x) is in lowest terms, we determine the asymptotes, symmetry, zeros, and y intercept if they exist. We graph the asymptotes as dashed lines, plot the zeros and y intercept, and plot several other points to determine how the graph approaches the asymptotes. Finally, we sketch the graph through the plotted points and approaching the asymptotes. EXAMPLE 21.3. (a) R(x) ¼ (a)

3 x2  1

Sketch a graph of each rational function R(x). (b) R(x) ¼

x2 4  x2 RðxÞ ¼

3 x2  1

has vertical asymptotes at x ¼ 1, and at x ¼ 1, a horizontal asymptote of y ¼ 0, and no holes. Since the numerator of R(x) is a constant, it does not have any zeros. Since R(0) ¼ 3, R(x) has a y intercept of (0, 23).

CHAP. 21]

237

RATIONAL FUNCTIONS

Plot the y intercept and graph the asymptotes as dashed lines. We determine some values of R(x) in each interval of the domain (1, 1), and (21, 1), and (1, 1). R(x) ¼ R(x), so R(x) is symmetric with respect to the y axis. R(2) ¼ R(2) ¼

3 ¼ 1, 22  1

R(0:5) ¼ R(0:5) ¼

3 ¼ 4 (0:5)2  1

Plot (2, 1), (2, 1), (0.5, 4), and (0:5, 4). Using the asymptotes as a boundary, we sketch the graph. The graph of R(x) ¼

3 x2  1

is shown in Fig. 21-1. R(x) ¼

(b)

x2 4  x2

has vertical asymptotes at x ¼ 2 and x ¼ 2, a horizontal asymptote of y ¼ 1, and no holes. The zeros of R(x) are for x ¼ 0. Since when x ¼ 0, R(0) ¼ 0 and (0, 0) is the zero and the y intercept. Plot the point (0, 0) and the vertical and horizontal asymptotes. We determine some values of R(x) in each interval of its domain (1, 2), (2, 2), and (2, 1). Since R(x) ¼ R(x), the graph is symmetric with respect to the y axis. R(3) ¼ R(3) ¼

32 9 9 ¼ ; ¼ 4  32 5 5

R(1) ¼ R(1) ¼

12 1 ¼ 4  12 3

Plot (3, 29/5), (23, 29/5), (1, 1/3), and (1, 21/3). Using the asymptotes as boundary lines, sketch the graph of R(x). The graph of R(x) ¼

x2 4  x2

is shown in Fig. 21-2.

Fig. 21-1

Fig. 21-2

238

21.5

RATIONAL FUNCTIONS

[CHAP. 21

GRAPHING RATIONAL FUNCTIONS USING A GRAPHING CALCULATOR

The graphing features of a graphing calculator allow for easy graphing of rational functions. However, unless the graphing calculator specifically plots the x values for the asymptotes, it connects the distinct branches of the rational function. You need to determine the vertical asymptotes, then set the scale on the x axis of the graphing window so that the values of the vertical asymptotes are used. Horizontal asymptotes must be read from the graph itself, since they are not drawn or labeled and only appear as a characteristic of the graph in the calculator display window. Holes are based on factors that can be canceled out from the rational function. These are difficult to locate from the display on a graphing calculator. In general, when using a graphing calculator to aid in the production of a graph on paper, it is good practice not to depend on the graphing calculator to determine the vertical asymptotes, horizontal asymptotes, or holes for a rational function. Determine these values for yourself and place them on the graph you are constructing. Use the display on the graphing calculator to indicate the location and shape of the graph and to guide your sketch of the graph.

Solved Problems 21.1

State the domain of each rational function R(x). (a) R(x) ¼

3x xþ2

(b) RðxÞ ¼

x3  2x2  3x x

(c) R(x) ¼

3x2  1 x3  x

SOLUTION (a) For R(x) ¼ 3x=(x þ 2), set x þ 2 ¼ 0 and see that R(x) is not defined for x ¼ 2: The domain of R(x) is {all real numbers except 2} or domain ¼ (1, 2) < (2, 1): (b) For R(x) ¼ (x3  2x2  3x)=x, we see that R(x) is not defined for x ¼ 0. Thus, the domain of R(x) is fall real numbers except 0g or domain ¼ (21, 0) < (0, 1). (c) For R(x) ¼ (3x2  1)=(x3  x), we set x3  x ¼ 0 and determine that for x ¼ 0, x ¼ 1, and x ¼ 1 R(x) is undefined. The domain of R(x) is fall real numbers except 1, 0, 1g or domain ¼ (1, 1) < (1, 0) < (0, 1) < (1, 1):

21.2

Determine the vertical asymptotes, horizontal asymptotes, and holes for each rational function R(x). (a) R(x) ¼

3x xþ2

(b) RðxÞ ¼

x3  2x2  3x x

(c) R(x) ¼

3x2  1 x3  x

SOLUTION (a) Values that make the denominator zero but do not make the numerator zero yield asymptotes. In R(x) ¼ 3x=(x þ 2), x ¼ 2 makes the denominator x þ 2 ¼ 0 but does not make the numerator 3x ¼ 0. Thus, x ¼ 2 is a vertical asymptote. Since the degree of the numerator 3x is 1 and the degree of the denominator x þ 2 is 1, R(x) has a horizontal asymptote of y ¼ 3/1 ¼ 3, where 3 is the lead coefficient of the numerator and 1 is the lead coefficient of the denominator. Since x ¼ 0 is the only value that makes the numerator 0 and x ¼ 0 does not make the denominator 0, R(x) has no holes in its graph. (b) Only x ¼ 0 makes the denominator of R(x) ¼ (x3  2x2  3x)=x zero. Since x ¼ 0 also makes the numerator zero, R(x) has no vertical asymptotes. Since the degree of the numerator of R(x) is 3 and the degree of the denominator is 1, the degree of the numerator exceeds the degree of the denominator and there is no horizontal asymptote.

CHAP. 21]

239

RATIONAL FUNCTIONS

Since x ¼ 0 makes both the numerator and denominator of R(x) zero, there is a hole in the graph of R(x) when x ¼ 0. We reduce R(x) to lowest terms and get R(x) ¼ x2  2x  3 when x = 0. The graph of this reduced form would have the value of 3 if x were 0, so the graph of R(x) has a hole at (0, 3). (c) Since x3  x ¼ 0 has solutions of x ¼ 0, x ¼ 1, and x ¼ 1, the vertical asymptotes of R(x) ¼ (3x2  1)= (x3  x) are x ¼ 0, x ¼ 1, and x ¼ 1: The degree of the numerator of R(x) is less than the degree of the denominator, so y ¼ 0 is the horizontal asymptote of R(x). The numerator is not zero for any values, x ¼ 1, x ¼ 0, and x ¼ 1, that make the denominator zero, so the graph of R(x) does not have any holes in it.

21.3

What are the zeros and y intercept of each rational function R(x)? (a) R(x) ¼

3x xþ2

(b) RðxÞ ¼

x3  2x2  3x x

(c) R(x) ¼

3x2  1 x3  x

SOLUTION (a) For R(x) ¼ 3x=(x þ 2), the numerator 3x is zero if x ¼ 0. Since x ¼ 0 does not make the denominator zero, there is a zero when x ¼ 0. Thus, (0, 0) is the zero of R(x). The y intercept is the value of y when x ¼ 0. Thus, (0, 0) is the point for the y intercept. (b) For R(x) ¼ (x3  2x2  3x)=x, the numerator x3  2x2  3x is zero when x ¼ 0, x ¼ 1, and x ¼ 3. However, x ¼ 0 makes the denominator zero, so it will not yield a zero of R(x). The zeros of R(x) are (3, 0) and (1, 0). From Problem 21.1(b), we know that x ¼ 0 is not in the domain of R(x). Thus, R(x) does not have a y intercept. pffiffiffi pffiffiffi 3 1)=(x  x), the numerator 3x2  1 ¼ 0 has solutions x ¼ 3=3 and x ¼  3=3. Thus the (c) For R(x) ¼ (3x2 p ffiffiffi pffiffiffi zeros of R(x) are ( 3=3; 0) and ( 3=3; 0): From Problem 21.1(c), we know that the domain of R(x) does not contain x ¼ 0, so R(x) does not have a y intercept.

21.4

Sketch a graph of each rational function R(x). (a) R(x) ¼

3x xþ2

(b) RðxÞ ¼

x3  2x2  3x x

(c) R(x) ¼

3x2  1 x3  x

SOLUTION From Problems 21.1, 21.2, and 21.3, we know the domains, vertical asymptotes, horizontal asymptotes, holes, zeros, and y intercepts for these three rational functions. We will use this information in graphing each of these rational functions. (a) R(x) ¼ 3x=(x þ 2) has a vertical asymptote of x ¼ 2, a horizontal asymptote of y ¼ 3, and (0, 0) is the point for the zero and y intercept. We draw dashed lines for the asymptotes and plot (0, 0). We select some x values and determine the points, then plot them. For x ¼ 4, 3, 1, and 2, we get the points (4, 6), (3, 9), (1, 3), and (2, 1.5). Now we sketch the graph of R(x) going through the plotted points and approaching the asymptotes. Since the domain of R(x) is separated into two parts, we will have two parts to the graph of R(x). See Fig. 21-3. (b) R(x) ¼ (x3  2x2  3x)=x ¼ x2  2x  3 when x= 0, and there is a hole at (0, 3). There are no asymptotes for the graph of R(x), there is no y intercept, but there are zeros at (3, 0) and (1, 0). We plot the zeros and place an open circle O around the point (0, 3) to indicate the hole in the graph. Now we select values of x, determine the corresponding points, and plot them. For x ¼ 2, 1, 2, and 4, we get the points (2, 5), (1, 4), (2, 3), and (4, 5). Since the domain of R(x) is separated into two parts by the x value for the hole of R(x), the graph of R(x) is separated into two parts by the hole at (0, 3). See Fig. 21-4. (c) R(x) ¼ (3x2  1)=(x3 p  ffiffix) of x ¼ 1, x ¼ 0, and x ¼ 1, a horizontal asymptote of ffi has verticalpasymptotes ffiffiffi y ¼ 0, and zeros of ( 3=3, 0) and ( 3=3, 0). We approximate the zeros to be (0.6, 0) and (0.6, 0) and we plot these points and graph the asymptotes. To be sure that we graph all the parts of R(x), we select x values from each interval in the domain. For x ¼ 2, 1:5, 0:75, 0:25, 0.25, 0.75, 1.5, and 2, we get

240

RATIONAL FUNCTIONS

Fig. 21-3

[CHAP. 21

Fig. 21-4

the points (2, 1:8), (1:5, 3:1), (0:75, 2:1), (0:25, 3:5), (0:25, 3:5), (0:75, 2:1), (1.5, 3.1), and (2, 1.8). Since the domain of R(x) is separated into four parts the graph of R(x) is in four separate parts. See Fig. 21-5.

Fig. 21-5

Supplementary Problems 21.5

State the domain of each rational function. 4 xþ2 1 (b) R(x) ¼  x2 x (c) R(x) ¼  2 x 4 (a) R(x) ¼

4 þx2 6x (e) R(x) ¼ xþ3 2x  5 ( f ) R(x) ¼ xþ4 (d ) R(x) ¼

x2

x3 þ 2 x2 x2 þ 4 (h) R(x) ¼ 27x3  3x x2 þ x (i) R(x) ¼ 2 x  5x þ 6

(g) R(x) ¼

CHAP. 21]

21.6

RATIONAL FUNCTIONS

Determine the asymptotes of each rational function. 4 x3 x (b) R(x) ¼ 2 x  16 3x þ 6 (c) R(x) ¼ x1 (a) R(x) ¼

21.7

3 xþ2

(b) R(x) ¼  (c) R(x) ¼

x 4

(d ) R(x) ¼

x3  27 x2

(g) R(x) ¼

x2  5x þ 6 x2 þ 6x þ 9

(e) R(x) ¼

x2 x3

(h) R(x) ¼

x3  1 x

( f ) R(x) ¼

x2  4 x2  1

(i) R(x) ¼

xþ3 x2 þ 2x þ 1

Graph each rational function. (a) R(x) ¼

2 x

(b) R(x) ¼

3 x2

2x  6 xþ1 xþ2 (e) R(x) ¼ x3

4  x2 x2  9 x (h) R(x) ¼  2 x 4

(d ) R(x) ¼

x2

1 1

( f ) R(x) ¼

x2

(g) R(x) ¼

x 9

(i) R(x) ¼

x2  5x þ 4 x2 þ 7x þ 6

Graph each rational function. (a) R(x) ¼

21.10

x2

2x þ 8 xþ3

(c) R(x) ¼  21.9

3 xþ4 2 (h) R(x) ¼ 2 x  7x þ 10 xþ5 (i) R(x) ¼ 5x (g) R(x) ¼ 

Determine the zeros and y intercept for each rational function. (a) R(x) ¼

21.8

x2  6x þ 9 x 2x2  5 (e) R(x) ¼ xþ2 x2 þ 2x ( f ) R(x) ¼ xþ3

(d ) R(x) ¼

xþ2 x2  4

(b) R(x) ¼

3x x2  2x

Graph each rational function. (a) R(x) ¼

2 x2 þ 4

(b) R(x) ¼

x2 þ 2 x3  x

ANSWERS TO SUPPLEMENTARY PROBLEMS (1, 2) < (2, 1) (1, 2) < (2, 1) (1, 2) < (2, 2) < (2, 1) (1, 2) < (2, 1) < (1, 1) (1, 3) < (3, 1)

21.5

(a) (b) (c) (d ) (e)

21.6

Vertical asymptotes (a) (b) (c) (d ) (e) ( f) (g) (h) (i)

x¼3 x ¼ 4, x ¼ 4 x¼1 x¼0 x ¼ 2 x ¼ 3 x ¼ 4 x ¼ 5, x ¼ 2 x¼5

(f ) (g) (h) (i)

(1, 4) < (4, 1) (1, 0) < (0, 1) (1, 1=3) < (1=3, 0) < (0, 1=3) < (1=3, 1) (1, 2) < (2, 3) < (3, 1)

Horizontal asymptotes y¼0 y¼0 y¼3 none none none y¼0 y¼0 y ¼ 1

241

242

21.7

21.8

RATIONAL FUNCTIONS

Zeros

y intercept

(a) (b) (c) (d ) (e) (f ) (g) (h) (i)

(0, 3/2) (0, 0) (0, 8/3) none (0, 0) (0, 4) (0, 2/3) none (0, 3)

none (0, 0) (4, 0) (3, 0) (0, 0) (2, 0), (2, 0) (3, 0), (2, 0) (1, 0) (3, 0)

(a) Fig. 21-6 (b) Fig. 21-7 (c) Fig. 21-8

(d ) Fig. 21-9 (e) Fig. 21-10 ( f ) Fig. 21-11

Fig. 21-6

Fig. 21-8

[CHAP. 21

(g) Fig. 21-12 (h) Fig. 21-13 (i) Fig. 21-14

Fig. 21-7

Fig. 21-9

CHAP. 21]

243

RATIONAL FUNCTIONS

Fig. 21-11

Fig. 21-10

Fig. 21-13

Fig. 21-12

Fig. 21-14

244

RATIONAL FUNCTIONS

Fig. 21-15

Fig. 21-17

[CHAP. 21

Fig. 21-16

Fig. 21-18

21.9

(a) Reduced to lowest terms R(x) ¼ 1=(x  2) when x = 2. Graph is Fig. 21-15. (b) Reduced to lowest terms R(x) ¼ 3=(x  2) when x = 0. Graph is Fig. 21-16.

21.10

(a) Fig. 21-17

(b) Fig. 21-18

CHAPTER 22

Sequences and Series

22.1

SEQUENCES

A sequence of numbers is a function defined on the set of positive integers. The numbers in the sequence are called terms. A series is the sum of the terms of a sequence.

22.2 A.

B.

ARITHMETIC SEQUENCES An arithmetic sequence is a sequence of numbers each of which, after the first, is obtained by adding to the preceding number a constant number called the common difference. Thus 3, 7, 11, 15, 19, . . . is an arithmetic sequence because each term is obtained by adding 4 to the preceding number. In the arithmetic sequence 50, 45, 40, . . . the common difference is 45 2 50 ¼ 40 2 45 ¼ 25. Formulas for arithmetic sequences (1) The nth term, or last term: l ¼ a þ (n  1)d n n (2) The sum of the first n terms: S ¼ (a þ l ) ¼ ½2a þ (n  1)d  2 2 where a ¼ first term of the sequence; d ¼ common difference; n ¼ number of terms; l ¼ nth term, or last term; S ¼ sum of first n terms. EXAMPLE 22.1. Consider the arithmetic sequence 3, 7, 11, . . . where a ¼ 3 and d ¼ 7  3 ¼ 11  7 ¼ 4: The sixth term is l ¼ a þ (n  1)d ¼ 3 þ (6  1)4 ¼ 23. The sum of the first six terms is n 6 S ¼ (a þ l ) ¼ (3 þ 23) ¼ 78 or 2 2

22.3 A.

n 6 S ¼ ½2a þ (n  1)d  ¼ ½2(3) þ (6  1)4 ¼ 78: 2 2

GEOMETRIC SEQUENCES A geometric sequence is a sequence of numbers each of which, after the first, is obtained by multiplying the preceding number by a constant number called the common ratio. 245

246

SEQUENCES AND SERIES

[CHAP. 22

Thus 5, 10, 20, 40, 80, . . . is a geometric sequence because each number is obtained by multiplying the preceding number by 2. In the geometric sequence 9, 23, 1  13 , 19 , . . . the common ratio is 3 1 1=3 1=9 1 ¼ ¼ ¼ ¼ : 9 3 1 1=3 3 B.

Formulas for geometric sequences. (1) The nth term, or last term: l ¼ ar n1 a(r n  1) rl  a ¼ ,r=1 (2) The sum of the first n terms: S ¼ r1 r1 where a ¼ first term; r ¼ common ratio; n ¼ number of terms; l ¼ nth term, or last term; S ¼ sum of first n terms. EXAMPLE 22.2.

Consider the geometric sequence 5, 10, 20, . . . where a ¼ 5 and r¼

10 20 ¼ ¼2 5 10

The seventh term is l ¼ ar n1 ¼ 5(271 ) ¼ 5(26 ) ¼ 320. The sum of the first seven terms is S¼

22.4

a(r n  1) 5(27  1) ¼ ¼ 635: r1 21

INFINITE GEOMETRIC SERIES

The sum to infinity (S1 ) of any geometric sequence in which the common ratio r is numerically less than 1 is given by S1 ¼ EXAMPLE 22.3.

a , 1r

where jrj , 1:

Consider the infinite geometric series 1 1 1 1  þ  þ  2 4 8

where a ¼ 1 and r ¼  12 : Its sum to infinity is S1 ¼

22.5

a 1 1 2 ¼ ¼ ¼ : 1  r 1  (1=2) 3=2 3

HARMONIC SEQUENCES

A harmonic sequence is a sequence of numbers whose reciprocals form an arithmetic sequence. Thus 1 1 1 1 1 , , , , ,... 2 4 6 8 10 is a harmonic sequence because 2, 4, 6, 8, 10, . . . is an arithmetic sequence.

22.6

MEANS

The terms between any two given terms of sequence are called the means between these two terms.

CHAP. 22]

247

SEQUENCES AND SERIES

Thus in the arithmetic sequence 3, 5, 7, 9, 11, . . . the arithmetic mean between 3 and 7 is 5, and four arithmetic means between 3 and 13 are 5, 7, 9, 11. In the geometric sequence 2, 24, 8, 216, . . . two geometric means between 2 and 216 are 24, 8. In the harmonic sequence 1 1 1 1 1 , , , , ,... 2 3 4 5 6 the harmonic mean between

1 1 1 1 1 1 1 1 and is , and three harmonic means between and are , , . 2 4 3 2 6 3 4 5

Solved Problems 22.1

22.2

Which of the following sequences are arithmetic sequences? (a) 1, 6, 11, 16, . . . 1 5 7 , 1, , , . . . (b) 3 3 3 (c) 4, 1, 6, 11, . . .

Yes, since 6 2 1 ¼ 11 2 6 ¼ 16 2 11 ¼ 5. (d ¼ 5) 1 5 7 5 2 Yes, since 1  ¼  1 ¼  ¼ : (d ¼ 23 ) 3 3 3 3 3 Yes, since 1  4 ¼ 6  (1) ¼ 11  (6) ¼ 5:

(d ) 9, 12, 16, . . . 1 1 1 , , ,... (e) 2 3 4 ( f ) 7, 9 þ 3p, 11 þ 6p, . . .

No, since 12  9 = 16  12. 1 1 1 1 No, since  =  . 3 2 4 3 Yes, with d ¼ 2 þ 3p.

Prove the formula S ¼ (n=2)(a þ l ) for the sum of the first n terms of an arithmetic sequence. SOLUTION The sum of the first n terms of an arithmetic sequence may be written

or

S ¼ a þ (a þ d ) þ (a þ 2d ) þ    þ l S ¼ l þ (l  d ) þ (l  2d ) þ    þ a

(n terms) (n terms)

where the sum is written in reversed order. Adding, Hence

22.3

2S ¼ (a þ l ) þ (a þ l ) þ (a þ l ) þ    þ (a þ l ) n 2S ¼ n(a þ l ) and S ¼ (a þ l ). 2

to n terms.

Find the 16th term of the arithmetic sequence: 4, 7, 10, . . . SOLUTION Here a ¼ 4, n ¼ 16, d ¼ 7  4 ¼ 10  7 ¼ 3, and l ¼ a þ (n  1)d ¼ 4 þ (16  1)3 ¼ 49.

22.4

(d ¼ 5)

Determine the sum of the first 12 terms of the arithmetic sequence: 3, 8, 13, . . . SOLUTION Here a ¼ 3, d ¼ 8  3 ¼ 13  8 ¼ 5, n ¼ 12, and n 12 S ¼ ½2a þ (n  1)d  ¼ ½2(3) þ (12  1)5 ¼ 366: 2 2 Otherwise: l ¼ a þ (n  1)d ¼ 3 þ (12  1)5 ¼ 58 and n 12 S ¼ (a þ l ) ¼ (3 þ 58) ¼ 366: 2 2

248

22.5

SEQUENCES AND SERIES

[CHAP. 22

Find the 40th term and the sum of the first 40 terms of the arithmetic sequence: 10, 8, 6, . . . SOLUTION Here d ¼ 8  10 ¼ 6  8 ¼ 2, a ¼ 10, n ¼ 40: Then l ¼ a þ (n  1)d ¼ 10 þ (40  1)(2) ¼ 68 n 40 S ¼ (a þ l ) ¼ (10  68) ¼ 1160: 2 2

22.6

and

Which term of the sequence 5, 14, 23, . . . is 239? SOLUTION l ¼ a þ (n  1)d,

22.7

239 ¼ 5 þ (n  1)9,

9n ¼ 243

and the required term is n ¼ 27.

Compute the sum of the first 100 positive integers exactly divisible by 7. SOLUTION The sequence is 7, 14, 21, . . . an arithmetic sequence in which a ¼ 7, d ¼ 7, n ¼ 100. Hence

22.8

n 100 S ¼ ½2a þ (n  1)d  ¼ ½2(7) þ (100  1)7 ¼ 35 350: 2 2

How many consecutive integers, beginning with 10, must be taken for their sum to equal 2035? SOLUTION The sequence is 10, 11, 12, . . . an arithmetic sequence in which a ¼ 10, d ¼ 1, S ¼ 2035. n S ¼ ½2a þ (n  1)d , 2 n n we obtain 2035 ¼ ½20 þ (n  1)1, 2035 ¼ (n þ 19), n2 þ 19n  4070 ¼ 0, 2 2 (n  55)(n þ 74) ¼ 0, n ¼ 55, 74: Hence 55 integers must be taken. Using

22.9

How long will it take to pay off a debt of $880 if $25 is paid the first month, $27 the second month, $29 the third month, etc.? SOLUTION n S ¼ ½2a þ (n  1)d , 2 n we obtain 880 ¼ ½2(25) þ (n  1)2, 880 ¼ 24n þ n2 , n2 þ 24n  880 ¼ 0, 2 (n  20)(n þ 44) ¼ 0, n ¼ 20, 44: The debt will be paid off in 20 months.

From

22.10 How many terms of the arithmetic sequence 24, 22, 20, . . . are needed to give the sum of 150? Write the terms. SOLUTION n 150 ¼ ½48 þ (n  1)(2), n2  25n þ 150 ¼ 0, (n  10)(n  15) ¼ 0, 2 For n ¼ 10: 24, 22, 20, 18, 16, 14, 12, 10, 8, 6. For n ¼ 15: 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0, 22, 24.

n ¼ 10, 15:

CHAP. 22]

SEQUENCES AND SERIES

249

22.11 Determine the arithmetic sequence whose sum to n terms is n2 þ 2n: SOLUTION The nth term ¼ sum to n terms 2 sum to n  1 terms ¼ n2 þ 2n  ½(n  1)2 þ 2(n  1) ¼ 2n þ 1. Thus the arithmetic sequence is 3, 5, 7, 9, . . .

22.12 Show that the sum of n consecutive odd integers beginning with 1 equals n2 . SOLUTION We are to find the sum of the arithmetic sequence 1, 3, 5, . . . to n terms. n n Then a ¼ 1, d ¼ 2, n ¼ n and S ¼ ½2a þ (n  1)d  ¼ ½2(1) þ (n  1)2 ¼ n2 : 2 2

22.13 Find three numbers in an arithmetic sequence such that the sum of the first and third is 12 and the product of the first and second is 24. SOLUTION Let the numbers in the arithmetic sequence be (a 2 d ), a, (a þ d ). Then (a 2 d ) þ (a þ d ) ¼ 12 or a ¼ 6. Since (a  d ) a ¼ 24, (6  d )6 ¼ 24 or d ¼ 2: Hence the numbers are 4, 6, 8.

22.14 Find three numbers in an arithmetic sequence whose sum is 21 and whose product is 280. SOLUTION Let the numbers be (a  d), a, (a þ d): Then (a  d) þ a þ (a þ d) ¼ 21 or a ¼ 7. Since (a  d )(a)(a þ d ) ¼ 280, a(a2  d 2 ) ¼ 7(49  d 2 ) ¼ 280 and d ¼ +3: The required numbers are 4, 7, 10 or 10, 7, 4.

22.15 Three numbers are in the ratio of 2 : 5 : 7. If 7 is subtracted from the second, the resulting numbers form an arithmetic sequence. Determine the original numbers. SOLUTION Let the original numbers be 2x, 5x, 7x. The resulting numbers in the arithmetic sequence are 2x, (5x – 7), 7x. Then (5x – 7) – 2x ¼ 7x – (5x – 7) or x ¼ 14. Hence the original numbers are 28, 70, 98.

22.16 Compute the sum of all integers between 100 and 800 that are divisible by 3. SOLUTION The arithmetic sequence is 102, 105, 108, . . . , 798. Then l ¼ a þ (n  1)d, 798 ¼ 102 þ (n  1)3, n ¼ 233, and n 233 S ¼ (a þ l) ¼ (102 þ 798) ¼ 104 850: 2 2

22.17 A slide of uniform grade is to be built on a level surface and is to have 10 supports equidistant from each other. The heights of the longest and shortest supports will be 4212 feet and 2 feet respectively. Determine the required height of each support. SOLUTION From l ¼ a þ (n – 1) d we have 4212 ¼ 2 þ (10  1) d and d ¼ 4 12 ft: Thus the heights are 2, 612 , 11, 1512 , 20, 2412 , 29, 3312 , 38, 4212 feet respectively.

250

SEQUENCES AND SERIES

[CHAP. 22

22.18 A freely falling body, starting from rest, falls 16 ft during the first second, 48 ft during the second second, 80 ft during the third second, etc. Calculate the distance it falls during the fifteenth second and the total distance it falls in 15 seconds from rest. SOLUTION Here d ¼ 48  16 ¼ 80  48 ¼ 32: During the 15th second it falls a distance l ¼ a þ (n  1) d ¼ 16 þ (15  1) 32 ¼ 464 ft: n 15 Total distance covered during 15 sec is S ¼ (a þ l) ¼ (16 þ 464) ¼ 3600 ft: 2 2

22.19 In a potato race, 8 potatoes are placed 6 ft apart on a straight line, the first being 6 ft from the basket. A contestant starts from the basket and puts one potato at a time into the basket. Find the total distance she must run in order to finish the race. SOLUTION n 8 Here a ¼ 2  6 ¼ 12 ft and l ¼ 2(6  8) ¼ 96 ft. Then S ¼ (a þ l) ¼ (12 þ 96) ¼ 432 ft. 2 2

22.20 Show that if the sides of a right triangle are in an arithmetic sequence, their ratio is 3 : 4 : 5. SOLUTION Let the sides be (a  d), a, (a þ d), where the hypotenuse is (a þ d). Then (a þ d)2 ¼ a2 þ (a  d)2 or a ¼ 4d: Hence (a  d) : a : (a þ d) ¼ 3d : 4d : 5d ¼ 3 : 4 : 5.

22.21 Derive the formula for the arithmetic mean (x) between two numbers p and q. SOLUTION Since p, x, q are in an arithmetic sequence, we have x  p ¼ q  x or x ¼ 12 ( p þ q).

22.22 Find the arithmetic mean between each of the following pairs of numbers. (a) 4 and 56. pffiffiffi pffiffiffi (b) 3 2 and 6 2. (c) a þ 5d and a  3d:

4 þ 56 ¼ 30: 2 pffiffiffi pffiffiffi pffiffiffi 3 2 þ (6 2) 3 2 : ¼ Arithmetic mean ¼ 2 2 (a þ 5d) þ (a  3d) Arithmetic mean ¼ ¼ a þ d: 2

Arithmetic mean ¼

22.23 Insert 5 arithmetic means between 8 and 26. SOLUTION We require an arithmetic sequence of the form 8, —, —, —, —, —, 26; thus a ¼ 8, l ¼ 26 and n ¼ 7. Then l ¼ a þ (n  1) d, 26 ¼ 8 þ (7  1) d, d ¼ 3: The five arithmetic means are 11, 14, 17, 20, 23.

22.24 Insert between 1 and 36 a number of arithmetic means so that the sum of the resulting arithmetic sequence will be 148.

CHAP. 22]

251

SEQUENCES AND SERIES

SOLUTION S ¼ 12 n(a þ l),

148 ¼ 12 n(1 þ 36),

l ¼ a þ (n  1) d,

37n ¼ 296 and n ¼ 8:

36 ¼ 1 þ (8  1) d,

7d ¼ 35

and

d ¼ 5:

The complete arithmetic sequence is 1, 6, 11, 16, 21, 26, 31, 36.

22.25 Which of the following sequences are geometric sequences? 6 12 ¼ ¼ 2: (r ¼ 2) 3 6   12 9 3 3 Yes, since ¼ ¼ : r¼ 6 12 4 4 3 9 ¼ ¼ 3: (r ¼ 3) Yes, since 1 3 4 9 No, since = . 1 4   1=3 2=9 2 2 Yes, since ¼ ¼ : r¼ 1=2 1=3 3 3   3 1=h 1=2h 1 1 ¼ ¼ 2: Yes, since r¼ 2 2h 2h 2h 1=h

(a) 3, 6, 12, . . .

Yes, since

(b) 16, 12, 9, . . . (c) 21, 3, 29, . . . (d) 1, 4, 9, . . . (e)

1 1 2 , , ,... 2 3 9

( f ) 2h,

1 1 , ,... h 2h3

22.26 Prove the formula S¼

a(r n  1) r1

for the sum of the first n terms of a geometric sequence. SOLUTION The sum of the first n terms of a geometric sequence may be written (1)

S ¼ a þ ar þ ar 2 þ ar 3 þ    þ ar nl

(n terms):

Multiplying (1) by r, we obtain (2)

rS ¼ ar þ ar 2 þ ar 3 þ    þ arn1 þ ar n

(n terms):

Subtracting (1) from (2), rS  S ¼ ar n  a,

(r  1)S ¼ a(r n  1)

and

S¼

a(r n  1) : r1

22.27 Find the 8th term and the sum of the first eight terms of the sequence 4, 8, 16, . . . SOLUTION Here a ¼ 4, r ¼ 8/4 ¼ 16/8 ¼ 2, n ¼ 8. The 8th term is l ¼ ar n1 ¼ 4(2)81 ¼ 4(27 ) ¼ 4(128) ¼ 512: The sum of the first eight terms is

S¼

a(r n  1) 4(28  1) 4(256  1) ¼ ¼ ¼ 1020: r1 21 1

252

SEQUENCES AND SERIES

[CHAP. 22

22.28 Find the 7th term and the sum of the first seven terms of the sequence 9, 26, 4, . . . SOLUTION a ¼ 9,

Here

Then the 7th term is S¼

l ¼ ar n1

6 4 2 ¼ ¼ : 9 6 3  71 2 64 ¼9  ¼ : 3 81 r¼

a(r n  1) a(1  r n ) 9½1  (2=3)7  9½1  (128=2187) 463 ¼ ¼ ¼ ¼ r1 1r 1  (2=3) 5=3 81

22.29 The second term of a geometric sequence is 3 and the fifth term is 81/8. Find the eighth term. SOLUTION 81 ar 4 81=8 27 ¼ , 2nd term ¼ ar ¼ 3: Then , r3 ¼ 8 3 8 ar   81 27 2187 ¼ . Hence the 8th term ¼ ar 7 ¼ (ar 4) r 3 ¼ 8 8 64 5th term ¼ ar 4 ¼

and

3 r¼ . 2

22.30 Find three numbers in a geometric sequence whose sum is 26 and whose product is 216. SOLUTION Let the numbers in geometric sequence be a/r, a, ar. Then (a/r)(a)(ar) = 216, a3 ¼ 216 and a ¼ 6. Also a/r þ a þ ar ¼ 26, 6/r þ 6 þ 6r ¼ 26, 6r 2 2 20r þ 6 ¼ 0 and r ¼ 1/3, 3. For r ¼ 1/3, the numbers are 18, 6, 2; for r ¼ 3, the numbers are 2, 6, 18.

22.31 The first term of a geometric sequence is 375 and the fourth term is 192. Find the common ratio and the sum of the first four terms. SOLUTION 1st term ¼ a ¼ 375, 4th term ¼ ar3 ¼ 192. Then 375r 3 ¼ 192, r 3 ¼ 64/125 and r ¼ 4/5. The sum of the first four terms is S¼

a(1  r n ) 375½1  (4=5)4  ¼ ¼ 1107: 1r 1  4=5

22.32 The first term of a geometric sequence is 160 and the common ratio is 3/2. How many consecutive terms must be taken to give a sum of 2110? SOLUTION a (r n  1) S¼ , 1r

160½(3=2)n  1 2110 ¼ , 3=2  1

 n 3 211 1 ¼ , 2 32

 n  5 3 243 3 ¼ , ¼ 2 32 2

n ¼ 5:

The five consecutive terms are 160, 240, 360, 540, 810.

22.33 In a geometric sequence consisting of four terms in which the ratio is positive, the sum of the first two terms is 8 and the sum of the last two terms is 72. Find the sequence.

CHAP. 22]

253

SEQUENCES AND SERIES

SOLUTION The four terms are a, ar, ar 2 , ar 3 : Then a þ ar ¼ 8 and ar 2 þ ar 3 ¼ 72. ar 2 þ ar 3 ar2 (1 þ r) 72 ¼ ¼ r2 ¼ ¼ 9, a þ ar a(1 þ r) 8

Hence

so that r ¼ 3:

Since a þ ar ¼ 8, a ¼ 2 and the sequence is 2, 6, 18, 54.

22.34 Prove that x, x þ 3, x þ 6 cannot be a geometric sequence. SOLUTION If x, x þ 3, x þ 6 is a geometric sequence then r¼

xþ3 xþ6 ¼ ; x xþ3

x2 þ 6x þ 9 ¼ x2 þ 6x

or

9 ¼ 0:

Since this equality can never be true, x, x þ 3, x þ 6 cannot be a geometric sequence.

22.35 A boy agrees to work at the rate of one cent the first day, two cents the second day, four cents the third day, eight cents the fourth day, etc. How much would he receive at the end of 12 days? SOLUTION Here a ¼ 1, r ¼ 2, n ¼ 12. S¼

a(r n  1) ¼ 212  1 ¼ 4096  1 ¼ 4095¢ ¼ $40:95: r1

22.36 It is estimated that the population of a certain town will increase 10% each year for four years. What is the percentage increase in population after four years? SOLUTION Let p denote the initial population. After one year the population is 1.10p, after two years (1.10)2p, after three years (1.10)3p, after four years (1.10)4p ¼ 1.46p. Thus the population increases 46%.

22.37 From a tank filled with 240 gallons of alcohol, 60 gallons are drawn off and the tank is filled up with water. Then 60 gallons of the mixture are removed and replaced with water, etc. How many gallons of alcohol remain in the tank after 5 drawings of 60 gallons each are made? SOLUTION After the first drawing, 240 2 60 ¼ 180 gal of alcohol remain in the tank. After the second drawing,     240  60 3 180 ¼ 180 gal 240 4 of alcohol remain, etc. The geometric sequence for the number of gallons of alcohol remaining in the tank after successive drawings is    2 3 3 3 , 180 180, 180 , . . . where a ¼ 180, r ¼ : 4 4 4

254

SEQUENCES AND SERIES

[CHAP. 22

After the fifth drawing (n ¼ 5):  4 3 l ¼ ar n1 ¼ 180 ¼ 57 gal 4 of alcohol remain.

22.38 A sum of $400 is invested today at 6% per year. To what amount will it accumulate in five years if interest is compounded (a) annually, (b) semiannually, (c) quarterly? SOLUTION Let P ¼ initial principal, i ¼ interest rate per period, S ¼ compound amount after n periods. At end of 1st period: interest ¼ Pi, new amount ¼ P þ Pi ¼ P(1 þ i). At end of 2nd period: interest ¼ P(l þ i)i, new amount ¼ P(l þ i) þ P(l þ i)i ¼ P(l þ i)2. Compound amount at end of n periods is S ¼ P(1 þ i)n. (a) Since there is 1 interest period per year, n ¼ 5 and i ¼ 0.06. S ¼ P(1 þ i)n ¼ 400(1 þ 0:06)5 ¼ 400(1:3382) ¼ $535:28 (b) Since there are 2 interest periods per year, n ¼ 2(5) ¼ 10 and i ¼ 12(0.06) ¼ 0.03. S ¼ P(1 þ i)n ¼ 400(1 þ 0:03)10 ¼ 400(1:3439) ¼ $537:56: (c) Since there are 4 interest periods per year, n ¼ 4(5) ¼ 20 and i ¼ 14(0.06) = 0.015. S ¼ P(1 þ i)n ¼ 400(1 þ 0:015)20 ¼ 400(1:3469) ¼ $538:76:

22.39 What sum (P) should be invested in a loan association at 4% per annum compounded semiannually, so that the compound amount (S) will be $500 at the end of 312 years? SOLUTION Since there are 2 interest periods per year, n ¼ 2(3 12 ) ¼ 7 (periods) and the interest rate per period is i ¼ 12(0.04) ¼ 0.02. Then S ¼ P(1 þ i)n or P ¼ S(1 þ i)n ¼ 500(1 þ 0:02)7 ¼ 500(0:870 56) ¼ $435:28:

22.40 Derive the formula for the geometric mean, G, between two numbers p and q. SOLUTION

pffiffiffiffiffi Since p, G, q are in geometric sequence, we have G/p ¼ q/G, G 2 ¼ pq and G ¼ + pq. pffiffiffiffiffi It is customary to take G ¼ pq if p and q are positive. pffiffiffiffiffi and G ¼  pq if p and q are negative.

22.41 Find the geometric mean between each of the following pairs of numbers. pffiffiffiffiffiffiffiffiffi (a) 4 and 9. G ¼ 4(9) ¼ 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (b) 22 and 28. G ¼  (2)(8) ¼ 4 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi (c) 7 þ 3 and 7  3. G ¼ ( 7 þ 3)( 7  3) ¼ 7  3 ¼ 2 22.42 Show that the arithmetic mean A of two positive numbers p and q is greater than or equal to their geometric mean G.

CHAP. 22]

SEQUENCES AND SERIES

255

SOLUTION

pffiffiffiffiffi Arithmetic mean of p and q is A ¼ 12 ð p þ qÞ. Geometric mean of p and q is G ¼ pq. p ffiffiffiffiffi p ffiffiffiffiffi p ffiffi ffi p ffiffi ffi Then A  G ¼ 12 ( p þ q)  pq ¼ 12 ( p  2 pq þ q) ¼ 12 ( p  q)2 . pffiffiffi pffiffiffi Now 12 ( p  q)2 is always positive or zero; hence A  G. (A ¼ G if and only if p ¼ q.)

22.43 Insert two geometric means between 686 and 2. SOLUTION We require a geometric sequence of the form 686, —, —, 2 where a ¼ 686, l ¼ 2, n ¼ 4. Then l ¼ ar n1 , 2 ¼ 686r 3 , r 3 ¼ 1=343 and r ¼ 1/7. Thus the geometric sequence is 686, 98, 14, 2 and the means are 98, 14. Note. Actually, r 3 ¼ 1=343 is satisfied by three different values of r, one of the roots being real and two imaginary. It is customary to exclude geometric sequences with imaginary numbers.

22.44 Insert five geometric means between 9 and 576. SOLUTION We require a geometric sequence of the form 9,—, —, —, —, —, 576 where a ¼ 9, l ¼ 576, n ¼ 7. Then l ¼ ar n1 , 576 ¼ 9r 6 , r 6 ¼ 64, r 3 ¼ +8 and r ¼ +2: Thus the sequences are 9, 18, 36, 72, 144, 288, 576 and 9, 218, 36, 272, 144, 2288, 576; and the corresponding means are 18, 36, 72, 144, 288 and 218, 36, 272, 144, 2288.

22.45 Find the sum of the infinite geometric series. 1 1 (a) 2 þ 1 þ þ þ    2 4 (b)

1 2 4 8  þ  þ  3 9 27 81

(c) 1 þ

1 1 þ  þ 1:04 (1:04)2

S1 ¼

a 2 ¼ ¼4 1  r 1  1=2

S1 ¼

a 1=3 1 ¼ ¼ 1  r 1  (2=3) 5

S1 ¼

a 1 1:04 104 ¼ ¼ ¼ ¼ 26 1  r 11=1:04 1:04  1 4

22.46 Express each of the following repeating decimals as a rational fraction. (a) 0.444 . . .

(b) 0.4272727 . . .

(c) 6.305305 . . .

(d ) 0.78367836 . . .

SOLUTION (a) 0.444. . . ¼ 0.4 þ 0.04 þ 0.004 þ . . . , where a ¼ 0.4, r ¼ 0.1. S1 ¼

a 0:4 0:4 4 ¼ ¼ ¼ 1  r 1  0:1 0:9 9

(b) 0.4272727. . . ¼ 0.4 þ 0.0272727 . . . 0.0272727 . . . ¼ 0.027 þ 0.00027 þ 0.0000027 þ . . . , where a ¼ 0.027, r ¼ 0.01. S1 ¼ 0:4 þ

a 0:027 27 4 3 47 ¼ 0:4 þ ¼ 0:4 þ ¼ þ ¼ 1r 1  0:01 990 10 110 110

(c) 6.305305. . . ¼ 6 þ 0.305305 . . . 0.305305 . . . ¼ 0.305 þ 0.000305 þ . . . , where a ¼ 0.305, r ¼ 0.001. S1 ¼ 6 þ

a 0:305 305 305 ¼6þ ¼6þ ¼6 1r 1  0:001 999 999

256

SEQUENCES AND SERIES

[CHAP. 22

(d ) 0.78367836 . . . ¼ 0.7836 þ 0.00007836 þ . . . , where a ¼ 0.7836, r ¼ 0.0001. S1 ¼

a 0:7836 7836 2612 ¼ ¼ ¼ 1  r 1  0:0001 9999 3333

22.47 The distances passed over by a certain pendulum bob in succeeding swings form the geometric sequence 16, 12, 9, . . . inches respectively. Calculate the total distance traversed by the bob before coming to rest. SOLUTION S1 ¼

a 16 16 ¼ ¼ ¼ 64 inches 1  r 1  3=4 1=4

1 1 1 22.48 Find the least number of terms of the series þ þ þ    that should be taken so that their sum will 3 6 12 differ from their sum to infinity by less than 1/1000. SOLUTION Let S1 = sum to infinity, Sn ¼ sum to n terms. Then S1  Sn ¼

a a(1  r n ) ar n :  ¼ 1r 1r 1r

It is required that ar n 1 , , where a ¼ 1=3, r ¼ 1=2: 1  r 1000 Then (1=3)(1=2)n 1 , , 1  1=2 1000

1 1 2 , 3(2n ) . 2000, 2n . 666 : , 3(2n ) 2000 3

When n ¼ 9, 2n , 666 23; when n ¼ 10, 2n . 666 23. Thus at least 10 terms should be taken.

22.49 Which of the following sequences are harmonic sequences? (a)

1 1 1 , , ,... 3 5 7

(b) 2, 4, 6, . . . (c)

1 2 1 , , ,... 12 15 3

is a harmonic sequence since 3, 5, 7, . . . is an arithmetic sequence. 1 1 1 , , , . . . is not an arithmetic sequence. 2 4 6 15 is a harmonic sequence since 12, , 3, . . . is an arithmetic sequence. 2

is not a harmonic sequence since

1 1 1 , .... 22.50 Compute the 15th term of the harmonic sequence , , 4 7 10 SOLUTION The corresponding arithmetic sequence is 4, 7,10, . . .; its 15th term is l ¼ a þ (n  1) d ¼ 4 þ (15  1)3 ¼ 46: 1 Hence the 15th term of the harmonic progression is . 46

22.51 Derive the formula for the harmonic mean, H, between two numbers p and q. SOLUTION 1 1 1 Since p, H, q is a harmonic sequence, , , is an arithmetic sequence. p H q

CHAP. 22]

257

SEQUENCES AND SERIES

Then

1 1 1 1  ¼  , H p q H

2 1 1 pþq ¼ þ ¼ H p q pq

and

H¼

2pq : pþq

Another method. 1 1 Harmonic mean between p and q ¼ reciprocal of the arithmetic mean between and . p q   1 1 1 1 1 pþq Arithmetic mean between and ¼ þ : ¼ p q 2 p q 2pq Hence the harmonic mean between p and q ¼

2pq . pþq

22.52 What is the harmonic mean between 3/8 and 4? SOLUTION

  8 1 1 8 1 35 ¼ : þ Arithmetic mean between and ¼ 3 4 2 3 4 24 3 Hence the harmonic mean between and 4 ¼ 24/35. 8 2pq 2(3=8)(4) 24 ¼ ¼ : Or, by formula, harmonic mean ¼ pþq 3=8 þ 4 35

22.53 Insert four harmonic means between 1/4 and 1/64. SOLUTION To insert four arithmetic means between 4 and 64: l ¼ a þ (n  1)d, 64 ¼ 4 þ (6  1)d, d ¼ 12: Thus the four arithmetic means between 4 and 64 are 16, 28, 40, 52. Hence the four harmonic means between

1 1 1 1 1 1 and are , , , . 4 64 16 28 40 52

22.54 Insert three harmonic means between 10 and 20. SOLUTION To insert three arithmetic means between

1 1 and . 10 20

l ¼ a þ (n  1)d, Thus the three arithmetic means between

1 1 ¼ þ (5  1)d, 20 10

d¼

1 : 80

1 1 7 6 5 and are , , : 10 20 80 80 80

Hence the three harmonic means between 10 and 20 are

80 40 , , 16. 7 3

22.55 Determine whether the sequence 21, 24, 2 is in arithmetic, geometric or harmonic sequence. SOLUTION Since 24 2 (21) = 2 2 (24), it is not in arithmetic sequence. 4 2 Since = , it is not in geometric sequence. 1 4   1 1 1 1 1 1 Since , , are in arithmetic sequence, i.e.,  (1) ¼  , the given sequence is in harmonic 1  4 2 4 2 4 sequence.

258

SEQUENCES AND SERIES

[CHAP. 22

Supplementary Problems 22.56

22.57

Find the nth term and the sum of the first n terms of each arithmetic sequence for the indicated value of n. (a) 1, 7, 13, . . . n ¼ 100

(c) 226, 224, 222, . . . n ¼ 40

(e) 3, 412 , 6, . . .

(b) 2, 512, 9, . . . n ¼ 23

(d ) 2, 6, 10, . . .

( f ) x  y, x, x þ y, . . . n ¼ 30

n ¼ 16

n ¼ 37

Find the sum of the first n terms of each arithmetic sequence. (a) 1, 2, 3, . . .

(b) 2, 8, 14, . . .

(c) 1 12 , 5, 8 12 , . . .

22.58

An arithmetic sequence has first term 4 and last term 34. If the sum of its terms is 247, find the number of terms and their common difference.

22.59

An arithmetic sequence consisting of 49 terms has last term 28. If the common difference of its terms is 1/2, find the first term and the sum of the terms.

22.60

Find the sum of all even integers between 17 and 99.

22.61

Find the sum of all integers between 84 and 719 which are exactly divisible by 5.

22.62

How many terms of the arithmetic sequence 3, 7, 11, . . . are needed to yield the sum 1275?

22.63

Find three numbers in an arithmetic sequence whose sum is 48 and such that the sum of their squares is 800.

22.64

A ball starting from rest rolls down an inclined plane and passes over 3 in. during the 1st second, 5 in. during the 2nd second, 7 in. during the 3rd second, etc. In what time from rest will it cover 120 inches?

22.65

If 1¢ is saved the 1st day, 2¢ the 2nd day, 3¢ the third day, etc., find the sum that will accumulate at the end of 365 days.

22.66

The sum of 40 terms of a certain arithmetic sequence is 430, while the sum of 60 terms is 945. Determine the nth term of the arithmetic sequence.

22.67

Find an arithmetic sequence which has the sum of its first n terms equal to 2n2 þ 3n.

22.68

Determine the arithmetic mean between (a) 15 and 41, (b) 216 and 23, (c) 2  5x þ 2y.

22.69

(a) (b) (c) (d )

22.70

Find the nth term and the sum of the first n terms of each geometric sequence for the indicated value of n.

Insert 4 arithmetic means between 9 and 24. Insert 2 arithmetic means between 21 and 11. Insert 3 arithmetic means between x þ 2y and x þ 10y. Insert between 5 and 26 a number of arithmetic means such that the sum of the resulting arithmetic sequence will be 124.

(a) 2, 3, 9/2, . . . n ¼ 5 (b) 6, 212, 24, . . . n ¼ 9 (c) 1, 1/2, 1/4, . . . n ¼ 10 22.71

(d ) 1, 3, 9, . . . n¼8 (e) p 8, ffiffi4, 2, . . . ffi pffiffiffi n ¼ 12 ( f ) 3, 3, 3 3, . . . n ¼ 8

Find the sum of the first n terms of each geometric sequence. (a) 1, 1/3, 1/9, . . .

22.72

pffiffiffi pffiffiffi 3 and 4 þ 3 3, (d) x  3y and

(b) 4/3, 2, 3, . . .

(c) 1, 22, 4, . . .

A geometric sequence has first term 3 and last term 48. If each term is twice the previous term, find the number of terms and the sum of the geometric sequence.

CHAP. 22]

22.73

SEQUENCES AND SERIES

259

Prove that the sum S of the terms of a geometric sequence in which the first term is a, the last term is l and the common ratio is r is given by S¼

rl  a : r1

22.74

In a geometric sequence the second term exceeds the first term by 4, and the sum of the second and third terms is 24. Show that there are two possible geometric sequences satisfying these conditions and find the sum of the first 5 terms of each geometric sequence.

22.75

In a geometric sequence consisting of four terms, in which the ratio is positive, the sum of the first two terms is 10 and the sum of the last two terms is 22 12. Find the sequence.

22.76

The first two terms of a geometric sequence are b/(1 þ c) and b/(1 þ c)2. Show that the sum of n terms of this sequence is given by the formula   1  (1 þ c)n S¼b c

22.77

Find the sum of the first n terms of the geometric sequence: a  2b, ab2  2b3 , ab4  2b5 , . . .

22.78

The third term of a geometric sequence is 6 and the fifth term is 81 times the first term. Write the first five terms of the sequence, assuming the terms are positive.

22.79

Find three numbers in a geometric sequence whose sum is 42 and whose product is 512.

22.80

The third term of a geometric sequence is 144 and the sixth term is 486. Find the sum of the first five terms of the geometric sequence.

22.81

A tank contains a salt water solution in which is dissolved 972 lb of salt. One third of the solution is drawn off and the tank is filled with pure water. After stirring so that the solution is uniform, one third of the mixture is again drawn off and the tank is again filled with water. If this process is performed four times, what weight of salt remains in the tank?

22.82

The sum of the first three terms of a geometric sequence is 26 and the sum of the first six terms is 728.What is the nth term of the geometric sequence?

22.83

The sum of three numbers in geometric sequence is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in an arithmetic sequence. Find the geometric sequence.

22.84

Determine the geometric mean between: (a) 2 and 18, (b) 4 and 6, (c) 24 and 216,

(d ) a þ b and 4a þ 4b.

22.85

(a) Insert two geometric means between 3pand ffiffiffi 192. (b) Insert four geometric means between 2 and 8. (c) The geometric mean of two numbers is 8. If one of the numbers is 6, find the other.

22.86

The first term of an arithmetic sequence is 2; and the first, third and eleventh terms are also the first three terms of a geometric sequence. Find the sum of the first eleven terms of the arithmetic sequence.

22.87

How many terms of the arithmetic sequence 9, 11, 13, . . . must be added in order that the sum should equal the sum of nine terms of the geometric sequence 3, 26, 12, 224, . . .?

22.88

In a set of four numbers, the first three are in a geometric sequence and the last three are in an arithmetic sequence with a common difference of 6. If the first number is the same as the fourth, find the four numbers.

260

SEQUENCES AND SERIES

22.89

Find two numbers whose difference is 32 and whose arithmetic mean exceeds the geometric mean by 4.

22.90

Find the sum of the infinite geometric series. (a) 3 þ 1 þ 1/3 þ . . . (c) 1 þ 1/ 22 þ 1/24 þ . . . (b) 4 þ 2 þ 1 þ . . . (d ) 6 2 2 þ 2/3    

[CHAP. 22

(e) 4 2 8/3 þ 16/9     ( f ) 1 þ 0.1 þ 0.01 þ . . .

22.91

The sum of the first two terms of a decreasing geometric series is 5/4 and the sum to infinity is 9/4. Write the first three terms of the geometric series.

22.92

The sum of an infinite number of terms of a decreasing geometric series is 3, and the sum of their squares is also 3. Write the first three terms of the series.

22.93

The successive distances traversed by a swinging pendulum bob are respectively 36, 24, 16, . . . in. Find the distance which the bob will travel before coming to rest.

22.94

Express each repeating decimal as a rational fraction. (a) 0.121212 . . . (c) 0.270270 . . . (e) 0.1363636 . . . (b) 0.090909 . . . (d ) 1.424242 . . . ( f ) 0.428571428571428 . . .

22.95

(a) Find the 8th term of the harmonic sequence 2/3, 1/2, 2/5, . . . (b) Find the 10th term of the harmonic sequence 5, 30/7, 15/4, . . . (c) What is the nth term of the harmonic sequence 10/3, 2, 10/7, . . .?

22.96

Find the harmonic mean between each pair of numbers. pffiffiffi pffiffiffi (a) 3 and 6 (b) 1/2 and 1/3 (c) 3 and 2

(d ) a þ b and a 2 b

22.97

(a) Insert two harmonic means between 5 and 10. (b) Insert four harmonic means between 3/2 and 3/7.

22.98

An object moves at uniform speed a from A to B and then travels at uniform speed b from B to A. Show that the average speed in making the round trip is 2ab/(a þ b), the harmonic mean between a and b. Calculate the average speed if a ¼ 30 ft/sec and b ¼ 60 ft/sec.

ANSWERS TO SUPPLEMENTARY PROBLEMS 22.56

(a) l ¼ 595, S ¼ 29 800

(c) l ¼ 52, S ¼ 520

(e) l ¼ 57, S ¼ 1110

(b) l ¼ 19, S ¼ 93112

(d ) l ¼ 62, S ¼ 512

( f ) l ¼ x þ 28y, S ¼ 30x þ 405y

n(n þ 1) 2

22.57

(a)

22.58

n ¼ 13, d ¼ 5/2

22.59

a ¼ 4, S ¼ 784

22.60

2378

22.61

50 800

22.62

25

22.63

12, 16, 20

22.64

10 sec

(b) n(3n 2 l)

(c)

n(7n  1) 4

CHAP. 22]

SEQUENCES AND SERIES

22.65

$667.95

22.66

nþ1 2

22.67

5, 9, 13, 17, . . .

22.68

(a) 28,

22.69

(a) 12, 15, 18, 21 (b) 3, 7

22.70

(a) l ¼ 81/8, S ¼ 211/8

(c) l ¼ 1/512, S ¼ 1023/512

(b) l ¼ 1536, S ¼ 1026

(d ) l ¼ 2187, S ¼ 3280

nth term ¼ 4n þ 1 (c) 3 þ

(b) 7/2,

 n 

3 1 1 2 3

pffiffiffi 3,

(d) 3x 2 y/2

(c) x þ 4y, x þ 6y, x þ 8y (d ) The arithmetic sequence is 5, 8, 11, 14, 17, 20, 23, 26

(b)

8 3

 n  3 1 2

1  (2)n 3

(c)

22.71

(a)

22.72

n ¼ 5, S ¼ 93

22.74

2, 6, 18, . . . and S ¼ 242; 4, 8, 16, . . . and S ¼ 124

22.75

4, 6, 9, 27/2

22.77

(a  2b) þ (b2n  1) b2  1

22.78

2/3, 2, 6, 18, 54

22.79

2, 8, 32

22.80

844

22.81

192 lb

22.82

2 . 3n21

22.83

2, 4, 8

22.84

(a) 6

22.85

(a) 12, 48

22.86

187 or 22

22.87

19

22.88

8, 24, 2, 8

22.89

18, 50

22.90

(a) 9/2

22.91

3/4, 1/2, 1/3

22.92

3/2, 3/4, 3/8

pffiffiffi (b) 2 6

(c) 2 8

pffiffiffi pffiffiffi (b) 2, 2 2, 4, 4 2

(b) 8

(c) 4/3

(e) l ¼ 1/256, S ¼ 4095/256 pffiffiffi ( f ) l ¼ 81, S ¼ 120 þ 40 3

(d ) 2a þ 2b (c) 32/3

(d ) 9/2

(e) 12/5

( f ) 10/9

261

262

SEQUENCES AND SERIES

22.93

108 in.

22.94

(a) 4/33

22.95

(a) 1/5

22.96

(a) 4

22.97

(a) 6, 15/2

22.98

40 ft/sec

(b) 1/11 (b) 2 (b) 2/5

(c) 10/37

10 2n þ 1 pffiffiffi pffiffiffi (c) 6 2  4 3

(d ) 47/33

(c)

(b) 1, 3/4, 3/5, 1/2

(d )

a2  b2 a

(e) 3/22

[CHAP. 22

( f ) 3/7

CHAPTER 23

Logarithms 23.1

DEFINITION OF A LOGARITHM

If bx ¼ N, where N is a positive number and b is a positive number different from 1, then the exponent x is the logarithm of N to the base b and is written x ¼ logb N. EXAMPLE 23.1. Write 32 ¼ 9 using logarithmic notation. Since 32 ¼ 9, then 2 is the logarithm of 9 to the base 3, i.e., 2 ¼ log3 9. EXAMPLE 23.2. Evaluate log2 8. log2 8 is that number x to which the base 2 must be raised in order to yield 8, i.e., 2x ¼ 8, x ¼ 3. Hence log2 8 ¼ 3.

Both bx ¼ N and x ¼ logb N are equivalent relationships; bx ¼ N is called the exponential form and x ¼ logb N the logarithmic form of the relationship. As a consequence, corresponding to laws of exponents there are laws of logarithms. 23.2 I.

LAWS OF LOGARITHMS The logarithm of the product of two positive numbers M and N is equal to the sum of the logarithms of the numbers, i.e., logb MN ¼ logb M þ logb N:

II.

The logarithm of the quotient of two positive numbers M and N is equal to the difference of the logarithms of the numbers, i.e., M logb ¼ logb M  logb N: N

III.

The logarithm of the pth power of a positive number M is equal to p mutiplied by the logarithm of the number, i.e., logb M p ¼ p logb M:

EXAMPLES 23.3. (a) (a) (b) (c) (d )

Apply the laws of logarithms to each expression. pffiffiffi 17 (c) log7 53 (b) log10 (d) log10 3 2 log2 3(5) 24 log2 3(5) ¼ log2 3 þ log2 5 17 log10 ¼ log10 17  log10 24 24 log7 53 ¼ 3 log7 5 pffiffiffi 1 log10 3 2 ¼ log10 21=3 ¼ log10 2 3

263

264

23.3

LOGARITHMS

[CHAP. 23

COMMON LOGARITHMS

The system of logarithms whose base is 10 is called the common logarithm system. When the base is omitted, it is understood that base 10 is to be used. Thus log 25 ¼ log10 25. Consider the following table. Number N Exponential form of N log N

0.0001

0.001

0.01

0.1

1

10

100

1000

10 000

1024

1023

1022

1021

100

101

102

103

104

24

23

22

21

0

1

2

3

4

It is obvious that 101.5377 will give some number greater than 10 (which is 101) but smaller than 100 (which is 102). Actually, 101.5377 ¼ 34.49; hence log 34.49 ¼ 1.5377. The digit before the decimal point is the characteristic of the log, and the decimal fraction part is the mantissa of the log. ln the above example, the characteristic is 1 and the mantissa is .5377. The mantissa of the log of a number is found in tables, ignoring the decimal point of the number. Each mantissa in the tables is understood to have a decimal point preceding it, and the mantissa is always considered positive. The characteristic is determined by inspection from the number itself according to the following rules. (1)

(2)

23.4

For a number greater than 1, the characteristic is positive and is one less than the number of digits before the decimal point. For example: Number

5297

348

900

34.8

60

5.764

3

Characteristic

3

2

2

1

1

0

0

For a number less than 1, the characteristic is negative and is one more than the number of zeros immediately following the decimal point. The negative sign of the characteristic is written in either of these two ways: (a) above the characteristic as 1, 2, etc.; (b) as 9 10, 8 10, etc. Thus the characteristic of 0.3485 is 1 or 9  10, of 0.0513 is 2 or 8 10, and of 0.0024 is 3 or 7  10: USING A COMMON LOGARITHM TABLE

To find the common logarithm of a positive number use the table of common logarithms in Appendix A. Suppose it is required to find the log of the number 728. ln the table of common logarithms glance down the N column to 72, then horizontally to the right to column 8 and note the figure 8621, which is the required mantissa. Since the characteristic is 2, log 728 ¼ 2.8621. (This means that 728 ¼ 102.8621.) The mantissa for log 72.8, for log 7.28, for log 0.728, for log 0.0728, etc., is .8621, but the characteristics differ. Thus: log 728 ¼ 2.8621 log 0.728 ¼ 1:8621 or 9.8621 2 10 log 72.8 ¼ 1.8621 log 0.0728 ¼ 2:8621 or 8.8621 2 10 log 7.28 ¼ 0.8621 log 0:007 28 ¼ 3:8621 or 7.8621 2 10 When the number contains four digits, interpolate using the method of proportional parts. EXAMPLE 23.4. Find log 4.638. The characteristic is 0. The mantissa is found as follows. Mantissa of log 4640 ¼ :6665 Mantissa of log 4630 ¼ :6656 Tabular difference ¼ :0009

CHAP. 23]

265

LOGARITHMS

.8  tabular difference ¼ .000 72 or .0007 to four decimal places. Mantissa of log 4638 ¼ .6656 þ .0007 ¼ .6663 to four digits. Hence log 4.638 ¼ 0.6663.

The mantissa for log 4638, for log 463:8, for log 46:38, etc., is .6663, but the characteristics differ. Thus: log log log log

4638 ¼ 3.6663 463.8 ¼ 2.6663 46.38 ¼ 1.6663 4.638 ¼ 0.6663

log 0.4638 log 0.046 38 log 0.004 638 log 0.000 463 8

¼ 1:6663 ¼ 2:6663 ¼ 3:6663 ¼ 4:6663

or or or or

9:6663  10 8:6663  10 7:6663  10 6:6663  10

The antilogarithm is the number corresponding to a given logarithm. “The antilog of 3” means “the number whose log is 3”; that number is obviously 1000. EXAMPLES 23.5.

Find the value of N.

(a) log N ¼ 1:9058

(b) log N ¼ 7:8657  10

(c) log N ¼ 9:3842  10:

(a) ln the table the mantissa .9058 corresponds to the number 805. Since the characteristic of log N is 1, the number must have two digits before the decimal point; thus N ¼ 80.5 (or antilog 1.9058 ¼ 80.5). (b) ln the table the mantissa .8657 corresponds to the number 734. Since the characteristic is 7 2 10, the number must have two zeros immediately following the decimal point; thus N ¼ 0.007 34 (or antilog 7:8657  10 ¼ 0:007 34). (c) Since the mantissa .3842 is not found in the tables, interpolation must be used. Mantissa of log 2430 ¼ :3856 Mantissa of log 2420 ¼ :3838 Tabular difference ¼ .0018

Given mantissa ¼ :3842 Next smaller mantissa ¼ :3838 Difference ¼ .0004

4 (2430  2420) ¼ 2422 to four digits, and N = 0.2422. Then 2420 þ 18

23.5

NATURAL LOGARITHMS

The system of logarithms whose base is the constant e is called the natural logarithm system. When we want to indicate the base of a logarithm is e we write ln. Thus, ln 25 ¼ loge 25. The exponential form of ln a ¼ b is eb ¼ a. The number e is an irrational number that has a decimal expansion e ¼ 2.718 281 828 450 45 . . . . 23.6

USING A NATURAL LOGARITHM TABLE

To find the natural logarithm of a positive number use the table of natural logarithms in Appendix B. To find the natural logarithm of a number from 1 to 10, such as 5.26, we go down the N column to 5.2, then across to the right to the column headed by .06 to get the value 1.6601. Thus, ln 5.26 ¼ 1.6601. This means that 5:26 ¼ e1:6601 . If we want to find the natural logarithm of a number greater than 10 or less than one, we write the number in scientific notation, apply the laws of logarithms, and use the natural logarithm table and the fact that ln 10 ¼ 2:3026: EXAMPLES 23.6. (a) 346

Find the natural logarithm of each number.

(b) 0.0217

(a) ln 346 ¼ ln(3:46  102 ) ¼ ln 3:46 þ ln 102 ¼ ln 3:46 þ 2 ln 10 ¼ 1:2413 þ 2(2:3026) ¼ 1:2413 þ 4:6052 ln 346 ¼ 5:8465

266

LOGARITHMS

[CHAP. 23

(b) ln 0.0217 ¼ ln(2:17  102 ) ¼ ln 2:17 þ ln 102 ¼ ln 2:17  2 ln 10 ¼ 0:7747  2(2:3026) ¼ 0:7747  4:6052 ln 0:0217 ¼ 3:8305

The value of ln 4.638 cannot be found directly from the natural logarithm table since it has four significant digits, but we can interpolate to find it. ln 4:640 ¼ 1:5347 ln 4:630 ¼ 1:5326 tabular difference ¼ 0:0021 0.8  tabular difference ¼ 0.8  0.0021 ¼ 0.001 68 or 0.0017 to four decimal places. Thus, ln 4.638 ¼ ln 4.630 þ 0.0017 ¼ 1.5326 þ 0.0017 ¼ 1.5343. The antilogarithm of a natural logarithm is the number that has the given logarithm. The procedure for finding the antilogarithm of a natural logarithm less than 0 or greater than 2.3026, requires us to add or subtract multiples of ln 10 ¼ 2.3026 to bring the natural logarithm into the range 0 to 2.3026 that can be found from the table in Appendix B. EXAMPLES 23.7. (a) ln N ¼ 2.1564

Find the value of N. (b) ln N ¼ 24.9705

(c) ln N ¼ 1.8869

(a) ln N ¼ 2.1564 is between 0 and 2.3026, so we look in the natural logarithm table for 2.1564. It is in the table, so we get N from the sum of the numbers that head the row and column for 1.1564. Thus, N ¼ antilogarithm 2.1564 ¼ 8.64. (b) Since ln N ¼ 24.9705 is less than 0, we must rewrite it as a number between 0 and 2.3026 minus a multiple of 2.3026 ¼ ln 10. Since if we add 3 times 2.3026 to 24.9705 we get a positive number between 0 and 2.3026, we will rewrite 24.9705 as 1.9373 2 3(2.3026). ln N ¼ 24.9705 ¼ 1.9373 2 3(2.3026) ¼ ln 6.94 2 3 ln 10 ¼ ln 6.94 þ ln 1023 ¼ ln (6.94  1023) ln N ¼ ln 0.006 94 N ¼ 0.006 94

Note: ln 6.94 ¼ 1.9373 and ln 10 ¼ 2.3026

(c) Since ln N ¼ 1.8869 is between 0 and 2.3026, we look for 1.8869 in the natural logarithm table, but it is not there. We will have to use interpolation to find N. ln 6.600 ¼ 1.8871 ln 6:590 ¼ 1:8856 tabular difference ¼ 0.0015 N ¼ 6:590 þ

23.7

ln N ¼ 1.8869 ln 6:590 ¼ 1:8856 difference ¼ 0.0013

13 (6:600  6:590) ¼ 6:590 þ 0:009 ¼ 6:599 15

FINDING LOGARITHMS USING A CALCULATOR

If the number we want to find the logarithm of a number with four or more significant digits, we can round the number to four significant digits and use the logarithm tables and interpolation or we can use a scientific or graphing calculator to find the logarithm for the given number. The use of the calculator will yield a more accurate result.

CHAP. 23]

267

LOGARITHMS

A scientific calculator can be used to find logarithms and antilogarithms to base 10 or base e. Scientific calculators have keys for the log and ln functions, and the inverse of these yields the antilogarithms. Much of the computation once done using logarithms can be done directly on a scientific calculator. The advantages of doing a problem on the calculator are that the numbers rarely have to be rounded and that the problem can be worked quickly and accurately.

Solved Problems 23.1

Express each of the following exponential forms in logarithmic form: 1 1 (e) 82=3 ¼ : (a) pq ¼ r, (b) 23 ¼ 8, (c) 42 ¼ 16, (d ) 32 ¼ , 9 4 SOLUTION (a) q ¼ logp r,

23.2

(b) 3 ¼ log2 8,

(c) 2 ¼ log4 16,

2 1 (e)  ¼ log8 3 4

Express each of the following logarithmic forms in exponential form: 1 (d ) loga a3 ¼ 3, (b) log2 64 ¼ 6, (c) log1=4 ¼ 2, (a) log5 25 ¼ 2; 16 SOLUTION (a) 52 ¼ 25,

23.3

1 (d ) 2 ¼ log3 ; 9

(b) 26 ¼ 64,

(c)

 2 1 1 ¼ , 4 16

(d ) a3 ¼ a3 ,

(e) logr 1 ¼ 0:

(e) r0 ¼ 1

Determine the value of each of the following. (a) log4 64. Let log4 64 ¼ x; then 4x ¼ 64 ¼ 43 and x ¼ 3. (b) log3 81. Let log3 81 ¼ x; then 3x ¼ 81 ¼ 34 and x ¼ 4.  x (c) log1=2 8. Let log1=2 8 ¼ x; then 12 ¼ 8, (21 )x ¼ 23 , 2x ¼ 23 and x ¼ 3: pffiffiffi pffiffiffi (d) log 3 10 ¼ x, 10x ¼ 3 10 ¼ 101=3 , x ¼ 1=3 pffiffiffi pffiffiffi (e) log5 125 5 ¼ x, 5x ¼ 125 5 ¼ 53  51=2 ¼ 57=2 , x ¼ 7=2

23.4

Solve each of the following equations. (a) log3 x ¼ 2, 32 ¼ x, x ¼ 9 3 1 (b) log4 y ¼  , 43=2 ¼ y, y ¼ 2 8 (c) logx 25 ¼ 2, x2 ¼ 25, x ¼ +5. Since bases are positive, the solution is x ¼ 5.  3=2 9 2 9 4 4 8 2=3 2=3 ¼ , y ¼ , y¼ ¼ (d) logy ¼  , y is the required solution. 4 3 4 9 9 27 (e) log(3x2 þ 2x  4) ¼ 0,

23.5

100 ¼ 3x2 þ 2x  4, 3x2 þ 2x  5 ¼ 0,

x ¼ 1, 5=3:

Prove the laws of logarithms. SOLUTION Let M ¼ bx and N ¼ by ; then x ¼ logb M and y ¼ logb N: I. Since MN ¼ bx  by ¼ bxþy , then logb MN ¼ x þ y ¼ logb M þ logb N: II. III.

Since

M bx M ¼ ¼ bxy , then logb ¼ x  y ¼ logb M  logb N: N by N

Since M p ¼ (bx )p ¼ b px , then logb M p ¼ px ¼ p logb M:

268

23.6

LOGARITHMS

[CHAP. 23

Express each of the following as an algebraic sum of logarithms, using the laws I, II, III. (a) logb UVW ¼ logb (UV)W ¼ logb UV þ logb W ¼ logb U þ logb V þ logb W UV ¼ logb UV  logb W ¼ logb U þ logb V  logb W (b) logb W XYZ ¼ log XYZ  log PQ ¼ log X þ log Y þ log Z  ( log P þ log Q) (c) log PQ ¼ log X þ log Y þ log Z  log P  log Q U2 (d ) log 3 ¼ log U 2  log V 3 ¼ 2 log U  3 log V V U2V 3  ¼ log U 2 V 3  log W 4 ¼ log U 2 þ log V 3  log W 4 (e) log W4 ¼ 2 log U þ 3 log V  4 log W U 1=2 1 2 ¼ log U 1=2  log V 2=3 ¼ log U  log V 2 3 Vp2=3 ffiffiffiffiffi 3=2 3 x 3 3 x ffiffiffiffiffi ¼ loge 3=4 ¼ loge x3=2  loge y3=4 ¼ loge x  loge y (g) loge p 4 3 2 4 y y  p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 3 1 2 log a  log b þ log c (h) log 4 a2 b3=4 c1=3 ¼ 4 4 3 1 3 1 ¼ log a  log b þ log c 2 16 12

( f ) log

23.7

Given that log 2 ¼ 0:3010, log 3 ¼ 0:4771, log 5 ¼ 0:6990, log 7 ¼ 0:8451 (all base 10) accurate to four decimal places, evaluate the following. (a) log 105 ¼ log (3  5  7) ¼ log 3 þ log 5 þ log 7 ¼ 0:4771 þ 0:6990 þ 0:8451 ¼ 2:0212 (b) log 108 ¼ log (22  33 ) ¼ 2 log 2 þ 3 log 3 ¼ 2(0:3010) þ 3(0:4771) ¼ 2:0333 pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi 2 (c) log 3 72 ¼ log 3 32  23 ¼ log (32=3  2) ¼ log 3 þ log 2 ¼ 0:6191 3 24 3  23 (d ) log 2:4 ¼ log ¼ log ¼ log 3 þ 3 log 2  log 10 10 10 ¼ 0:4771 þ 3(0:3010)  1 ¼ 0:3801 81 (e) log 0:0081 ¼ log 4 ¼ log 81  log 104 ¼ log 34  log 104 10 ¼ 4 log 3  4 log 10 ¼ 4(0:4771)  4 ¼ 2:0916 or 7:9084  10 Note. ln exponential form this means 102:0916 ¼ 0:0081:

23.8

Express each of the following as a single logarithm (base is 10 unless otherwise indicated). 2 2 10 (a) log 2  log 3 þ log 5 ¼ log þ log 5 ¼ log (5) ¼ log 3 3 3 3 2 8 (b) 3 log 2  4 log 3 ¼ log 23  log 34 ¼ log 4 ¼ log 3 81 1 1 2 (c) log 25  log 64 þ log 27 ¼ log 251=2  log 641=3 þ log 272=3 2 3 3 5 5 45 ¼ log 5  log 4 þ log 9 ¼ log þ log 9 ¼ log (9) ¼ log 4 4 4 5 1 (d ) log 5  1 ¼ log 5  log 10 ¼ log ¼ log 10 2

CHAP. 23]

269

LOGARITHMS

(e) 2 log 3 þ 4 log 2  3 ¼ log 32 þ log 24  3 log 10 ¼ log 9 þ log 16  log 103 9  16 ¼ log (9  16)  log 103 ¼ log ¼ log 0:144 103 1 ( f ) 3 loga b  loga c ¼ loga b3 þ loga c1=2 ¼ loga (b3 c1=2 ) 2 23.9

ln each of the following equations, solve for the indicated letter in terms of the other quantities. (a) log2 x ¼ y þ c : x: (b) loga ¼ 2 log b : a: (c) loge I ¼ loge I0  t : I:

x ¼ 2yþc log a ¼ log b2 , a ¼ b2 loge I ¼ loge I0  t loge e ¼ loge I0 þ loge et ¼ loge I0 et , I ¼ I0 et

(d) 2 log x þ 3 log y ¼ 4 log z  2 : y: Solving for log y, 3 log y ¼ 4 log z  2  2 log x and 4 2 2 log y ¼ log z   log x ¼ log z4=3 þ log 102=3 þ log x2=3 ¼ log z4=3 102=3 x2=3 : 3 3 3 Hence y ¼ 102=3 x2=3 z4=3 : (e) log (x þ 3) ¼ log x þ log 3 : x:

log (x þ 3) ¼ log 3x,

x þ 3 ¼ 3x,

x ¼ 3=2

23.10 Determine the characteristic of the common logarithm of each of the following numbers. (a) 57 (b) 57.4

(c) 5.63 (e) 982.5 (g) 186 000 (d ) 35.63 ( f ) 7824 (h) 0.71

(i) 0.7314 ( j) 0.0325

(k) 0.0071 (l ) 0.0003

SOLUTION (a) 1 (b) 1

(c) 0 (d) 1

(e) 2 ( f) 3

(i) 9 2 10 ( j) 8 2 10

(g) 5 (h) 9 2 10

(k) 7 2 10 (l) 6 2 10

23.11 Verify each of the following common logarithms. (a) (b) (c) (d) (e) (f ) (g)

log 87.2 ¼ 1.9405 log 37 300 ¼ 4.5717 log 753 ¼ 2.8768 log 9.21 ¼ 0.9643 log 0.382 ¼ 9.5821 2 10 log 0.00 159 ¼ 7.2014 2 10 log 0.0256 ¼ 8.4082 2 10

(h) (i) ( j) (k) (l) (m) (n)

log 6.753 log 183.2 log 43.15 log 876 400 log 0.2548 log 0.043 72 log 0.009 848

¼ 0.8295 (8293 þ 2) ¼ 2.2630 (2625 þ 5) ¼ 1.6350 (6345 þ 5) ¼ 5.9427 (9425 þ 2) ¼ 9.4062 2 10 (4048 þ 14) ¼ 8.6407 2 10 (6405 þ 2) ¼ 7.9933 2 10 (9930 þ 3)

23.12 Verify each of the following. (a) Antilog 3.8531 ¼ 7130 (b) Antilog 1.4997 ¼ 31.6 (c) Antilog 9.8267 2 10 ¼ 0.671 (d) Antilog 7.7443 2 10 ¼ 0.005 55 (e) Antilog 0.1875 ¼ 1.54 ( f ) Antilog 2:3927 ¼ 0.0247 (g) Antilog 4.9360 ¼ 86 300

(h) (i) ( j) (k) (l) (m) (n)

Antilog 2.6715 ¼ 469.3 (3/9  10 ¼ 3 approx.) Antilog 4.1853 ¼ 15 320 (6/28  10 ¼ 2 approx.) Antilog 0.9245 ¼ 8.404 (2/5  10 ¼ 4) ¼ 0.4064 (4/11  10 ¼ 4 approx.) Antilog 1:6089 Antilog 8.8907 2 10 ¼ 0.077 75 (3/6  10 ¼ 5) Antilog 1.2000 ¼ 15.85 (13/27  10 ¼ 5 approx.) Antilog 7.2409 2 10 ¼ 0.001742 (4/25  10 ¼ 2 approx.)

270

LOGARITHMS

23.13 Write each of the following numbers as a power of 10: (a) 893, (b) 0.358. SOLUTION (a) We require x such that 10x ¼ 893. Then x ¼ log 893 ¼ 2:9509 and 893 ¼ 102:9509 . (b) We require x such that 10x ¼ 0:358. Then x ¼ log 0:358 ¼ 9:5539  10 ¼ 0:4461 and 0:358 ¼ 100:4461 .

Calculate each of the following using logarithms. 23.14 P ¼ 3:81  43:4 SOLUTION log P ¼ log 3:81 þ log 43:4 log 3:81 ¼ 0:5809 (þ) log 43:4 ¼ 1:6375 log P ¼ 2:2184 Hence P ¼ antilog 2.2184 ¼ 165.3. Note the exponential significance of the computation. Thus 3:81  43:4 ¼ 100:5809  101:6375 ¼ 100:5809þ1:6375 ¼ 102:2184 ¼ 165:3

23.15 P ¼ 73.42  0.004 62  0.5143 SOLUTION log P ¼ log 73: 42 þ log 0:004 62 þ log 0:5143 log 73: 42 ¼ 1:8658 (þ) log 0:004 62 ¼ 7:6646  10 (þ) log 0:5143 ¼ 9:7112  10 log P ¼ 19:2416  20 ¼ 9:2416  10: Hence P ¼ 0.1744.

23.16 P ¼

784:6  0:0431 28:23

SOLUTION log P ¼ log 784:6 þ log 0:0431  log 28:23 log 784:6 ¼ 2:8947 (þ) log 0:0431 ¼ 8:6345  10 11:5292  10 () log 28:23 ¼ 1:4507 log P ¼ 10:0785  10 ¼ 0:0785 P ¼ 1:198

23.17 P ¼ (7:284)5 SOLUTION log P ¼ 5 log 7:284 ¼ 5(0:8623) ¼ 4:3115 and P ¼ 20 490:

[CHAP. 23

CHAP. 23]

23.18 P ¼

271

LOGARITHMS

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 0:8532

SOLUTION 1 1 1 log P ¼ log 0:8532 ¼ (9:9310  10) ¼ (49:9310  50) ¼ 9:9862  10 and 5 5 5

P ¼ 0:9687:

pffiffiffiffiffiffiffiffiffiffiffi (78:41)3 142:3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffi p 23.19 P ¼ 4 0:1562 SOLUTION

1 1 log P ¼ 3 log 78:41 þ log 142:3  log 0:1562: 2 4 Numerator N 3 log 78:41 ¼ 3(1:8944) ¼ 5:6832 (þ) 12 log 142:3 ¼ 12 (2:1532) ¼ 1:0766

Denominator D ¼ 14 (9:1937  10) ¼ 14 (39:1937  40)

1 4 log 0:1562

log N ¼ 6:7598 ¼ 16:7598  10 () log D ¼ 9:7984  10

log D ¼ 9:7984  10

log P ¼ 6:9614 P ¼ 9 150 000

or

9:15  106

pffiffiffiffiffiffiffi 23.20 The period T of a simple pendulum of length l is given by the formula T ¼ 2p l=g, where g is the acceleration due to gravity. Find T (in seconds) if l ¼ 281.3 cm and g ¼ 981.0 cm/sec2. Take 2p ¼ 6:283. SOLUTION T ¼ 2p

rffiffiffiffiffiffiffiffiffiffiffi rffiffiffi l 281:3 ¼ 6:283 g 981:0

log T ¼ log 6:283 þ 12 ( log 281:3  log 981:0) log 6:283 ¼ ¼ 0:7982 (þ) 12 log 281:3 ¼ 12 (2:4492) ¼ 1:2246 2:0228 () 12 log 981:0 ¼ 12 (2:9917) ¼ 1:4959 log T ¼ 0:5269 T ¼ 3:365 seconds

23.21 Solve for x: 52xþ2 ¼ 35x1 . SOLUTION Taking logarithms, (2x þ 2) log 5 ¼ (5x  1) log 3: Then 2x log 5  5x log 3 ¼  log 3  2 log 5, x(2 log 5  5 log 3) ¼  log 3  2 log 5, and

x¼

log 3 þ 2 log 5 0:4771 þ 2(0:6990) 1:8751 ¼ ¼ : 5 log 3  2 log 5 5(0:4771)  2(0:6990) 0:9875 log 1:875 ¼ 10:2730  10 () log 0:9875 ¼ 9:9946  10 log

x ¼ 0:2784 x ¼ 1:898

272

LOGARITHMS

23.22 Find the value of each of these natural logarithms. (a) ln 5.78 (b) ln 8.62

(c) ln 3.456 (d ) ln 4.643

(e) ln 190 ( f ) ln 0.0084

(g) ln 2839 (h) ln 0.014 85

SOLUTION (a) ln 5:78 ¼ 1:7544

from natural logarithm table

(b) ln 8:62 ¼ 2:1541

from natural logarithm table

(c) ln 3:456 ¼ ln 3:45 þ 0:6( ln 3:46  ln 3:45) ¼ 1:2384 þ 0:6(1:2413  1:2384) ¼ 1:2384 þ 0:6(0:0029) ¼ 1:2384 þ 0:0017 ln 3:456 ¼ 1:2401 (d ) ln 4:643 ¼ ln 4:64 þ 0:3( ln 4:65  ln 4:64) ¼ 1:5347 þ 0:3(1:5369  1:5347) ¼ 1:5347 þ 0:3(0:0022) ¼ 1:5347 þ 0:0007 ln 4:643 ¼ 1:5354 (e) ln 190 ¼ ln (1:90  102 ) ¼ ln 1:90 þ ln 102 ¼ ln 1:90 þ 2 ln 10 ¼ 0:6419 þ 2(2:3026) ¼ 0:6419 þ 4:6052 ln 190 ¼ 5:2471 ( f ) ln 0:0084 ¼ ln (8:40  103 ) ¼ ln 8:40 þ ln 103 ¼ ln 8:40  3 ln 10 ¼ 2:1282  3(2:3026) ¼ 2:1282  6:9078 ln 0:0084 ¼ 4:7796 (g) ln 2839 ¼ ln (2:839  103 ) ¼ ln 2:839 þ ln 103 ¼ ½ln 2:83 þ 0:9(ln 2:84  ln 2:83) þ 3 ln 10 ¼ ½1:0403 þ 0:9(1:0438  1:0403) þ 3(2:3026) ¼ ½1:0403 þ 0:9(0:0035) þ 6:9078 ¼ ½1:0403 þ 0:0032 þ 6:9078 ¼ 1:0435 þ 6:9078 ln 2839 ¼ 7:9513 (h) ln 0:014 85 ¼ ln (1:485  102 ) ¼ ln 1:485 þ ln 102 ¼ ½ln 1:48 þ 0:5( ln 1:49  ln 1:48)  2 ln 10 ¼ ½0:3920 þ 0:5(0:3988  0:3920)  2 (2:3026) ¼ ½0:3920 þ 0:5(0:0068)  4:6052 ¼ ½0:3920 þ 0:0034  4:6052 ¼ 0:3954  4:6052 ln 0:014 85 ¼ 4:2098

23.23 Find the value of N. (a) ln N ¼ 2:4146

(b) ln N ¼ 0:9847

(c) ln N ¼ 1:7654

[CHAP. 23

CHAP. 23]

273

LOGARITHMS

SOLUTION (a) ln N ¼ 2:4146 ¼ 0:1120 þ 2:3026   0:1120  0:1044 ¼ ln 1:11 þ (1:12  1:11) þ ln 10 0:1133  0:1044   0:0076 (0:01) þ ln 10 ¼ ln 1:11 þ 0:0089 ¼ ln (1:11 þ 0:009) þ ln 10 ¼ ln 1:119 þ ln 10 ¼ ln (1:119  10) ln N ¼ ln 11:19 N ¼ 11:19 (b) ln N ¼ 0:9847   0:9847  0:9821 (2:68  2:67) ¼ ln 2:67 þ 0:9858  0:9821   0:0026 ¼ ln 2:67 þ (0:01) 0:0037 ¼ ln (2:67 þ 0:007) ln N ¼ ln 2:677 N ¼ 2:677 (c) ln N ¼ 1:7654 ¼ 0:5372  2:3026   0:5372  0:5365 ¼ ln 1:71 þ (1:72  1:71)  ln 10 0:5423  0:5365   0:0007 ¼ ln 1:71 þ (0:01) þ ln 101 0:0058 ¼ ln (1:71 þ 0:001) þ ln 101 ¼ ln 1:711 þ ln 101 ¼ ln (1:711  101 ) ln N ¼ ln 0:1711 N ¼ 0:1711

Supplementary Problems (a) log2 32,

Evaluate:

23.25

Solve each equation for the unknown. (a) log2 x ¼ 3 (b) log y ¼ 2

23.26

23.27

(b) log

ffiffiffiffiffi p 4 10,

23.24

(c) logx 8 ¼ 3 (d ) log3 (2x þ 1) ¼ 1

(c) log 3 1=9,

(d ) log 1=4 16,

(e) log4 x3 ¼ 3=2 ( f ) log(x1) (4x  4) ¼ 2

Express as an algebraic sum of logarithms. rffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi U3V 2 2x3 y (c) ln 3 x1=2 y1=2 (b) log (a) log 5 7 W z

(d ) log

xy3=2 z3 a2 b4

Solve each equation for the indicated letter in terms of the other quantities. (a) 2 log x ¼ log 16; x

(c) log3 F ¼ log3 4  2 log3 x; F

(b) 3 log y þ 2 log 2 ¼ log 32; y

(d ) ln (30  U) ¼ ln 30  2t; U

(e) loge ex , ( f ) log8 4.

274

LOGARITHMS

[CHAP. 23

23.28

Prove that if a and b are positive and =1, (loga b)(logb a) ¼ 1.

23.29

Prove that 10log N ¼ N where N . 0:

23.30

Determine the characteristic of the common logarithm of each number. (a) 248 (b) 2.48 (c) 0.024

23.31

23.32

(d ) 0.162 (e) 0.0006 ( f ) 18.36

(g) 1.06 (h) 6000 (i) 4

Find the common logarithm of (a) 237 (d ) 0.263 (g) (b) 28.7 (e) 0.086 (h) (c) 1.26 ( f ) 0.007 (i)

( j) 40.60 (k) 237.63 (l) 146.203

each number. 10 400 0.00 607 0.000 000 728

(m) 7 000 000 (n) 0.000 007

( j) 6 000 000 (k) 23.70 (l) 6.03

Find the antilogarithm of each of the following. (a) 2.8802 (c) 0.6946 (e) 8.3160 2 10 (b) 1.6590 (d ) 2:9042 ( f ) 7.8549 2 10

(g) 4.6618 (h) 0.4216

(m) 1 (n) 1000

(i) 1:9484 ( j) 9.8344 2 10

23.33

Find the common logarithm of each number by interpolation. (a) 1463 (c) 86.27 (e) 0.6041 (g) 1.006 (i) 460.3 (b) 810.6 (d ) 8.106 ( f ) 0.046 22 (h) 300.6 ( j) 0.003 001

23.34

Find the antilogarithm of each of the following by interpolation. (a) 2.9060 (c) 1.6600 (e) 3.7045 (g) 2.2500 (b) 1:4860 (d ) 1:9840 ( f ) 8:9266  10 (h) 0.8003

23.35

Write each number as a power of 10:

23.36

Evaluate. (a) (42.8)(3.26)(8.10)

(b)

(0:148)(47:6) 284

(b) 0.005 278.

5608 (e) (0:4536)(11 000) (f)

(3:92)3 (72:16) p ffiffiffiffiffiffiffiffi 4 654

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (g) 3:14 11:65=32

(1:86)(86:7) (c) (2:87)(1:88) (d )

(a) 45.4,

(i) 1:4700 ( j) 1.2925

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 906 (h) (3:142)(14:6) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (1600)(310:6)2 (i) (7290) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 (5:52)(2610) ( j) (7:36)(3:142)

2453 (67:2)(8:55)

23.37

Solve the following hydraulics equation:

23.38

Solve for x. (a) 3x ¼ 243 (b) 5x ¼ 1=125

(c) 2xþ2 ¼ 64 (d ) x2 ¼ 16

  20:0 0:0613 1:32 : ¼ 14:7 x

(e) x3=4 ¼ 8 ( f ) x2=3 ¼ 1=9

23.39

Solve each exponential equation:

(a) 42x1 ¼ 5xþ2 ,

23.40

Find the natural logarithms. (a) ln 2.367 (b) ln 8.532

(c) ln 4875

(g) 7x1=2 ¼ 4 (h) 3x ¼ 1 (b) 3x1 ¼ 4  513x :

(d) ln 0.000 189 4

(i) 5x2 ¼ 1 ( j) 22xþ3 ¼ 1

CHAP. 23]

23.41

275

LOGARITHMS

Find N, the antilogarithm of the given number. (a) ln N ¼ 0.7642 (b) ln N ¼ 1.8540

(c) ln N ¼ 8.4731

(d ) ln N ¼ 26.2691

ANSWERS TO SUPPLEMENTARY PROBLEMS 23.24

(a) 5

(b) 1/4

(c) 22

(d ) 22

(e) x

( f ) 2/3

23.25

(a) 8

(b) 0.01

(c) 1/2

(d ) 1

(e) 2

(f ) 5

23.26

(a) 3 log U þ 2 log V  5 log W (b)

1 1 ln x  ln y 6 6 3 (d ) log x  log y þ 3 log z  2 log a þ 4 log b 2 (c)

1 3 1 7 log 2 þ log x þ log y  log z 2 2 2 2 (c) F ¼ 4/x 2

(d) U ¼ 30(1 2 e 22t)

23.27

(a) 4

(b) 2

23.30

(a) 2 (b) 0

(c) 2 (d) 1

23.31

(a) 2.3747 (b) 1.4579 (c) 0.1004

(d ) 1:4200 (e) 2:9345 ( f ) 7:8451  10

23.32

(a) 759 (b) 45.6

(c) 4.95 (d ) 0.0802

(e) 0.0207 ( f ) 0.007 16

23.33

(a) 3.1653 (b) 2.9088

(c) 1.9359 (d ) 0.9088

(e) 1:7811 ( f ) 8:6648  10

23.34

(a) 805.4 (b) 0.3062

(c) 45.71 (d ) 0.9638

(e) 5064 ( f ) 0.084 45

23.35

(a) 101:6571

(b) 102:2776

23.36

(a) 1130 (b) 0.0248

(c) 29.9 (d ) 4.27

23.37

0.0486

23.38

(a) 5 (b) 23

23.39

(a) 3.958

23.40

(a) 0.8616

(b) 2.1438

(c) 8.4919

(d ) 28.5717

23.41

(a) 2.147

(b) 6.385

(c) 4784

(d ) 0.001 894

(e) 4 (f) 1

(c) 4 (d ) +1/4

(g) 0 (h) 3

(i) 0 ( j) 1

(k) 2 (l) 2

(g) 4.0170 (h) 3:7832 (i) 7:8621

(m) 6 (n) 6

( j) 6.7782 (k) 1.3747 (l) 0.7803 (g) 45 900 (h) 2.64

(m) 0.0000 (n) 3.0000

(i) 0.888 ( j) 0.683

(g) 0.0026 (h) 2.4780

(g) 177.8 (h) 6.314

(i) 0.2951 ( j) 19.61

(e) 1.124 ( f ) 860

(g) 1.90 (h) 4.44

(i) 145.5 ( j) 8.54

(e) 1/16 ( f ) +27

(g) 49/16 (h) 0

(i) 2 ( j) 23/2

(b) 0.6907

(i) 2.6631 ( j) 7:4773  10

CHAPTER 24

Applications of Logarithms and Exponents 24.1

INTRODUCTION

Logarithms have their major use in solving exponential equations and solving equations in which the variables are related logarithmically. To solve equations in which the variable is in the exponent, we generally start by changing the expression from exponential form to logarithmic form. 24.2

SIMPLE INTEREST

Interest is money paid for the use of a sum of money called the principal. The interest is usually paid at the ends of specified equal time intervals, such as monthly, quarterly, semiannually, or annually. The sum of the principal and the interest is called the amount. The simple interest, I, on the principal, P, for a time in years, t, at an interest rate per year, r, is given by the formula I ¼ Prt, and the amount, A, is found by A ¼ P þ Prt or A ¼ P(1 þ rt). EXAMPLE 24.1. paid on the loan?

If an individual borrows $800 at 8% per year for two and one-half years, how much interest must be I ¼ Prt I ¼ $800(0:08)(2:5) I ¼ $160

EXAMPLE 24.2. If a person invests $3000 at 6% per year for five years, how much will the investment be worth at the end of the five years? A ¼ P þ Prt A ¼ $3000 þ $3000(0:06)(5) A ¼ $3000 þ $900 A ¼ $3900

276

CHAP. 24]

24.3

APPLICATIONS OF LOGARITHMS AND EXPONENTS

277

COMPOUND INTEREST

Compound interest means that the interest is paid periodically over the term of the loan which results in a new principal at the end of each interval of time. If a principal, P, is invested for t years at an annual interest rate, r, compounded n times per year, then the amount, A, or ending balance is given by:  r nt A¼P 1þ n EXAMPLE 24.3.

Find the amount of an investment if $20 000 is invested at 6% compounded monthly for three years.  r nt A¼P 1þ n   0:06 12(3) A ¼ 20 000 1 þ 12 A ¼ 20 000(1 þ 0:005)36 A ¼ 20 000(1:005)36 log A ¼ log 20 000(1:005)36 log A ¼ log 20 000 þ 36 log 1:005 log A ¼ 4:3010 þ 36(0:002 15) log A ¼ 4:3010 þ 0:0774 log A ¼ 4:3784 A ¼ antilog 4:3784 A ¼ 2:39  104

log 2:39 ¼ 0:3784 and log 104 ¼ 4

A ¼ $23 900

When the interest is compounded more and more frequently, we get to a situation of continuously compounded interest. If a principal, P, is invested for t years at an annual interest rate, r, compounded continuously, then the amount, A, or ending balance, is given by: A ¼ Pert EXAMPLE 24.4.

Find the amount of an investment if $20 000 is invested at 6% compounded continuously for three years. A ¼ Pert A ¼ 20 000e0:06(3) A ¼ 20 000e0:18 ln A ¼ ln 20 000e0:18 ln A ¼ ln 20 000 þ ln e0:18 ln A ¼ ln(2:00  104 ) þ 0:18 ln e ln A ¼ ln 2:00 þ 4 ln 10 þ 0:18(1)

ln e ¼ 1

ln A ¼ 0:6931 þ 4(2:3026) þ 0:18

ln 2:00 ¼ 0:6931 and ln 10 ¼ 2:3026

ln A ¼ 10:0835 ln A ¼ 0:8731 þ 4(2:3026)

278

APPLICATIONS OF LOGARITHMS AND EXPONENTS

[CHAP. 24

  0:8731  0:8713 (2:40  2:39) þ 4 ln 10 ln A ¼ ln 2:39 þ 0:8755  0:8713   0:0018 ln A ¼ ln 2:39 þ (0:01) þ ln 104 0:0042 ln A ¼ ln(2:39 þ 0:004) þ ln 104 ln A ¼ ln 2:394 þ ln 104 ln A ¼ ln(2:394  104 ) ln A ¼ ln 23 940 A ¼ $23 940

In doing Examples 24.3 and 24.4 we found the answers to four significant digits. However, using the logarithm tables and doing interpolation results in some error. Also, we may have a problem if the interest is compounded daily, because when we divide r by n the result could be zero when rounded to thousandths. To deal with this problem and to get greater accuracy, we can use five-place logarithm tables, calculators, or computers. Generally, banks and other businesses use computers or calculators to get the accuracy they need. EXAMPLE 24.5. Use a scientific or graphing calculator to find the amount of an investment if $20 000 is invested at 6% compounded monthly for three years.  r nt A¼P 1þ n   0:06 12(3) A ¼ $20 000 1 þ 12 A ¼ $20 000(1:005)36

use the power key to compute (1:005)36

A ¼ $23 933:61 To the nearest cent, the amount has been increased by $33.61 from the amount found in Example 24.3. It is possible to compute the answer to the nearest cent here, while we were able to compute the result to the nearest ten dollars in Example 24.3. EXAMPLE 24.6. Use a scientific or graphing calculator to find the amount of an investment if $20 000 is invested at 6% compounded continuously for three years. A ¼ Pert A ¼ $20 000e0:06(3) A ¼ $20 000e0:18

use the inverse of ln x to compute e0:18

A ¼ $23 944:35 To the nearest cent, the amount has increased by $4.35 from the amount found in Example 24.4. The greater accuracy was possible because the calculator computes with more decimal places in each operation and then the answer is rounded. In our examples, we rounded to hundredths because cents are the smallest units of money that have a general usefulness. Most calculators compute with 8, 10, or 12 significant digits in doing the operations.

24.4

APPLICATIONS OF LOGARITHMS

The loudness, L, of a sound (in decibels) perceived by the human ear depends on the ratio of the intensity, I, of the sound to the threshold, I0 , of hearing for the average human ear.   I L ¼ 10 log I0

CHAP. 24]

EXAMPLE 24.7. human ear.

APPLICATIONS OF LOGARITHMS AND EXPONENTS

279

Find the loudness of a sound that has an intensity 10 000 times the threshold of hearing for the average   I I0   10 000I0 L ¼ 10 log I0

L ¼ 10 log

L ¼ 10 log 10 000 L ¼ 10 (4) L ¼ 40 decibels

Chemists use the hydrogen potential, pH, of a solution to measure its acidity or basicity. The pH of distilled water is about 7. If the pH of a solution exceeds 7, it is called a base, but if its pH is less than 7 it is called an acid. If ½Hþ  is the concentration of hydrogen ions in moles per liter, the pH is given by the formula: pH ¼  log ½Hþ  EXAMPLE 24.8.

Find the pH of the solution whose concentration of hydrogen ions is 5:32  105 moles per liter. pH ¼  log ½Hþ  pH ¼  log (5:32  105 ) pH ¼ ½log 5:32 þ log 105  pH ¼  log 5:32  (5) log 10 pH ¼  log 5:32 þ 5(1) pH ¼ 0:7259 þ 5 pH ¼ 4:2741

log 10 ¼ 1

pH ¼ 4:3

Seismologists use the Richter scale to measure and report the magnitude of earthquakes. The magnitude or Richter number of an earthquake depends on the ratio of the intensity, I, of an earthquake to the reference intensity, I0 , which is the smallest earth movement that can be recorded on a seismograph. Richter numbers are usually rounded to the nearest tenth or hundredth. The Richter number is given by the formula: R ¼ log

  I I0

EXAMPLE 24.9. If the intensity of an earthquake is determined to be 50 000 times the reference intensity, what is its reading on the Richter scale?   I R ¼ log I0   50 000I0 R ¼ log I0 R ¼ log 50 000 R ¼ 4:6990 R ¼ 4:70

280

24.5

APPLICATIONS OF LOGARITHMS AND EXPONENTS

[CHAP. 24

APPLICATIONS OF EXPONENTS

The number e is involved in many functions occurring in nature. The growth curve of many materials can be described by the exponential growth equation: A ¼ A0 ert where A0 is the initial amount of the material, r is the annual rate of growth, t is the time in years, and A is the amount of the material at the ending time. EXAMPLE 24.10. The population of a country was 2 400 000 in 1990 and it has an annual growth rate of 3%. If the growth is exponential, what will its population be in 2000? A ¼ A0 ert A ¼ 2 400 000e(0:03)(10) A ¼ 2 400 000e0:3 A ¼ 2 400 000(1:350) A ¼ 3 240 000

N ln N ln N N

¼ e0:3 ¼ 0:3 ln e ¼ 0:3 ¼ 1:350

The decay or decline equation is similar to the growth except the exponent is negative. A ¼ A0 ert where A0 is the initial amount, r is the annual rate of decay or decline, t is the time in years, and A is the ending amount. EXAMPLE 24.11. A piece of wood is found to contain 100 grams of carbon-14 when it is removed from a tree. If the rate of decay of carbon-14 is 0.0124% per year, how much carbon-14 will be left in the wood after 200 years? A ¼ A0 ert A ¼ 100e0:000 124 (200) A ¼ 100e0:0248 A ¼ 100(0:9755) A ¼ 97:55 grams

N ln N ln N ln N ln N ln N N

¼ e0:0248 ¼ 0:0248 ¼ ln 2:2778  2:3026 ¼ ln 9:755  ln 10 ¼ ln (9:755  101 ) ¼ ln 0:9755 ¼ 0:9755

Solved Problems 24.1

A woman borrows $400 for 2 years at a simple interest rate of 3%. Find the amount required to repay the loan at the end of 2 years. SOLUTION Interest I ¼ Prt ¼ 400(0.03)(2) ¼ $24. Amount A ¼ principal P þ interest I ¼ $424.

24.2

Find the interest I and amount A for (a) $600 for 8 months (2/3 yr) at 4%, (b) $1562.60 for 3 years, 4 months (10/3 yr) at 312 %. SOLUTION (a) I ¼ Prt ¼ 600(0.04)(2/3) ¼ $16.

A ¼ P þ I ¼ $616.

(b) I ¼ Prt ¼ 1562.60(0.035)(10/3) ¼ $182.30.

A ¼ P þ I ¼ $1744.90.

CHAP. 24]

24.3

281

APPLICATIONS OF LOGARITHMS AND EXPONENTS

What principal invested at 4% for 5 years will amount to $1200? SOLUTION A ¼ P(1 þ rt)

P¼

or

A 1200 1200 ¼ ¼ ¼ $1000: 1 þ rt 1 þ (0:04)(5) 1:2

The principal of $1000 is called the present value of $1200. This means that $1200 to be paid 5 years from now is worth $1000 now (the interest rate being 4%).

24.4

What rate of interest will yield $1000 on a principal of $800 in 5 years? SOLUTION A ¼ P(1 þ rt)

24.5

or

r¼

A  P 1000  800 ¼ ¼ 0:05 or 5%: Pt 800(5)

A man wishes to borrow $200. He goes to the bank where he is told that the interest rate is 5%, interest payable in advance, and that the $200 is to be paid back at the end of one year. What interest rate is he actually paying? SOLUTION The simple interest on $200 for 1 year at 5% is I ¼ 200(0:05)(1) ¼ $10. Thus he receives $200 2 $10 ¼ $190. Since he must pay back $200 after a year, P ¼ $190, A ¼ $200, t ¼ 1 year. Thus r¼

A  P 200  190 ¼ ¼ 0:0526, Pt 190(1)

i.e., the effective interest rate is 5.26%.

24.6

A merchant borrows $4000 under the condition that she pay at the end of every 3 months $200 on the principal plus the simple interest of 6% on the principal outstanding at the time. Find the total amount she must pay. SOLUTION Since $4000 is to be paid (excluding interest) at the rate of $200 every 3 months, it will take 4000/200(4) ¼ 5 years, i.e., 20 payments. Interest paid at 1st payment ( for first 3 months)

¼ 4000(0:06)

Interest paid at 2nd payment

¼ 3800(0:06)

Interest paid at 3rd payment .. .

¼ 3600(0:06) .. .

Interest paid at 20th payment

¼ 200(0:06)

1 41 41 4

1 4

¼ $60:00: ¼ $57:00: ¼ $54:00: .. . ¼ $3:00:

The total interest is 60 þ 57 þ 54 þ    þ 9 þ 6 þ 3: an arithmetic sequence with the sum given by S ¼ (n=2)(a þ l), where a ¼ 1st term, l ¼ last term, n ¼ number of terms. Then S ¼ (20/2)(60 þ 3) ¼ $630, and the total amount she must pay is $4630.

24.7

What will $500 deposited in a bank amount to in 2 years if interest is compounded semiannually at 2%?

282

APPLICATIONS OF LOGARITHMS AND EXPONENTS

[CHAP. 24

SOLUTION Method 1. Without formula. At end of 1st half year, interest ¼ 500(0:02)(12) At end of 2nd half year, interest ¼ 505(0:02)(12) At end of 3rd half year, interest ¼ At end of 4th half year, interest ¼

510:05(0:02)(12) 515:15(0:02)(12)

¼ $5:00 ¼ $5:05: ¼ $5:10: ¼ $5:15:

Total interest ¼ $20:30:

Total amount ¼ $520:30:

Method 2. Using formula. P ¼ $500, i ¼ rate per period ¼ 0.02/2 ¼ 0.01. n ¼ number of periods ¼ 4. A ¼ P(1 þ i)n ¼ 500(1:01)4 ¼ 500(1:0406) ¼ $520:30: Note. (1:01)4 may be evaluated by the binomial formula, logarithms or tables.

24.8

Find the compound interest and amount of $2800 in 8 years at 5% compounded quarterly. SOLUTION A ¼ P(1 þ i)n ¼ 2800(1 þ 0:05=4)32 ¼ 2800(1:0125)32 ¼ 2800(1:4881) ¼ $4166:68: Interest ¼ A  P ¼ $4166:68  $2800 ¼ $1366:68:

24.9

What rate of interest compounded annually is the same as the rate of interest of 6% compounded semiannually? SOLUTION Amount from principal P in 1 year at rate r ¼ P(1 þ r). Amount from principal P in 1 year at rate 6% compounded semiannually = P(1 þ 0:03)2  The amounts are equal if P(1 þ r) ¼ P(1.03)2, 1 þ r ¼ (1.03)2, r ¼ 0.0609 or 6.09%. The rate of interest i per year compounded a given number of times per year is called the nominal rate. The rate of interest r which, if compounded annually, would result in the same amount of interest is called the effective rate. In this example, 6% compounded semiannually is the nominal rate and 6.09% is the effective rate.

24.10 Derive a formula for the effective rate in terms of the nominal rate. SOLUTION Let r ¼ effective interest rate, i ¼ interest rate per annum compounded k times per year, i.e. nominal rate. Amount from principal P in 1 year at rate r ¼ P(1 þ r). Amount from principal P in 1 year at rate i compounded k times per year = P(1 þ i=k)k . The amounts are equal if P(1 þ r) ¼ P(1 þ i=k)k . Hence r ¼ (1 þ i=k)k  1.

24.11 The population in a country grows at a rate of 4% compounded annually. At this rate, how long will it take the population to double?

CHAP. 24]

SOLUTION

APPLICATIONS OF LOGARITHMS AND EXPONENTS

283

 r nt A¼P 1þ n 2P ¼ P(1 þ 0:04)t 2 ¼ (1:04)t log 2 ¼ t log (1:04) t¼

log 2 log 1:04

t¼

0:3010 0:0170

n¼1

t ¼ 17:7 years

24.12 If $1000 is invested at 10% compounded continuously, how long will it take the investment to triple? SOLUTION A ¼ Pert 3000 ¼ 1000e0:10t 3 ¼ e0:10t ln 3 ¼ 0:10t

ln e ¼ 1

ln 3 0:10 1:0986 t¼ 0:10 t¼

t ¼ 10:986 t ¼ 11:0

24.13 Find the pH of blood if the concentration of hydrogen ions is 3:98  108 . SOLUTION pH ¼  log½Hþ  pH ¼  log (3:98  108 ) pH ¼  log 3:98  (8) log 10 pH ¼ 0:5999 þ 8 pH ¼ 7:4001 pH ¼ 7:40

24.14 An earthquake in San Francisco in 1989 was reported to have a Richter number of 6.90. How does the intensity of the earthquake compare with the reference intensity? SOLUTION I I0 I 6:90 ¼ log I0 R ¼ log

log

I ¼ 6:90 I0

284

APPLICATIONS OF LOGARITHMS AND EXPONENTS

I ¼ antilog 6:90 I0

antilog 0:9000 ¼ 7:94 þ ¼ 7:94 þ

I ¼ (antilog 6:90)I0 I ¼ (7:943  106 )I0 I ¼ 7 943 000I0

[CHAP. 24

0:9000  0:8998 (0:01) 0:9004  0:8998 0:0002 (0:01) 0:0006

¼ 7:94 þ 0:003 antilog 0:9000 ¼ 7:943 antilog 6:9000 ¼ 7:943  106

24.15 The population of the world increased from 2.5 billion in 1950 to 5.0 billion in 1987. If the growth was exponential, what was the annual growth rate? SOLUTION A ¼ A0 ert 5:0 ¼ 2:5er(37) 2 ¼ e37r ln 2 ¼ 37r ln e 0:6931 ¼ 37r(1) 0:6931 ¼ 37r 0:01873 ¼ r r ¼ 0:0187 r ¼ 1:87%

24.16 In Nigeria the rate of deforestation is 5.25% per year. If the decrease in forests in Nigeria is exponential, how long will it take until only 25% of the current forests are left? SOLUTION A ¼ A0 ert 0:25A0 ¼ A0 e0:0525t 0:25 ¼ e0:0525t ln 0:25 ¼ 0:0525t ln e ln (2:5  101 ) ¼ 0:0525t(1) ln 2:5  ln 10 ¼ 0:0525t 0:9163  2:3026 ¼ 0:0525t  1:3863 ¼ 0:0525t 26:405 ¼ t t ¼ 26:41 years

Supplementary Problems 24.17

If $5.13 interest is earned in two years on a deposit of $95, then what is the simple annual interest rate?

24.18

If $500 is borrowed for one month and $525 must be paid back at the end of the month, what is the simple annual interest?

CHAP. 24]

APPLICATIONS OF LOGARITHMS AND EXPONENTS

285

24.19

If $4000 is invested at a bank that pays 8% interest compounded quarterly, how much will the investment be worth in 6 years?

24.20

If $8000 is invested in an account that pays 12% interest compounded monthly, how much will the investment be worth after 10 years?

24.21

A bank tried to attract new, large, long-term investments by paying 9.75% interest compounded continuously if at least $30 000 was invested for at least 5 years. If $30 000 is invested for 5 years at this bank, how much would the investment be worth at the end of the 5 years?

24.22

What interest will be earned if $8000 is invested for 4 years at 10% compounded semiannually?

24.23

What interest will be earned if $3500 is invested for 5 years at 8% compounded quarterly?

24.24

What interest will be earned if $4000 is invested for 6 years at 8% compounded continuously?

24.25

Find the amount that will result if $9000 is invested for 2 years at 12% compounded monthly.

24.26

Find the amount that will result if $9000 is invested for 2 years at 12% compounded continuously.

24 27

In 1990 an earthquake in Iran was said to have about 6 times the intensity of the 1989 San Francisco earthquake, which had a Richter number of 6.90. What is the Richter number of the Iranian earthquake?

24.28

Find the Richter number of an earthquake if its intensity is 3 160 000 times as great as the reference intensity.

24.29

An earthquake in Alaska in 1964 measured 8.50 on the Richter scale. What is the intensity of this earthquake compared with the reference intensity?

24.30

Find the intensity as compared with the reference intensity of the 1906 San Francisco earthquake if it has a Richter number of 8.25.

24.31

Find the Richter number of an earthquake with an intensity 20 000 times greater than the reference intensity.

24.32

Find the pH of each substance with the given concentration of hydrogen ions.

24.33

(a) beer: ½Hþ  ¼ 6:31  105

(c) vinegar: ½Hþ  ¼ 6:3  103

(b) orange juice: ½Hþ  ¼ 1:99  104

(d ) tomato juice: ½Hþ  ¼ 7:94  105

Find the approximate hydrogen ions concentration, ½Hþ , for the substances with the given pH. (a) apples: pH ¼ 3:0

(b) eggs: pH ¼ 7:8

24.34

If gastric juices in your stomach have a hydrogen ion concentration of 1:01  101 moles per liter, what is the pH of the gastric juices?

24.35

A relatively quiet room has a background noise level of 32 decibels. How many times the hearing threshold intensity is the intensity of a relatively quiet room?

24.36

If the intensity of an argument is about 3 980 000 times the hearing threshold intensity, what is the decibel level of the argument?

24.37

The population of the world compounds continuously. If in 1987 the growth rate was 1.63% annually and an initial population of 5 billion people, what will the world population be in the year 2000?

286

APPLICATIONS OF LOGARITHMS AND EXPONENTS

[CHAP. 24

24.38

During the Black Plague the world population declined by about 1 million from 4.7 million to about 3.7 million during the 50-year period from 1350 to 1400. If world population decline was exponential, what was the annual rate of decline?

24.39

If the world population grew exponentially from 1.6 billion in 1900 to 5.0 billion in 1987, what was the annual rate of population growth?

24.40

If the deforestation of El Salvador continues at the current rate for 20 more years only 53% of the present forests will be left. If the decline of the forests is exponential, what is the annual rate of deforestation for El Salvador?

25.41

A bone is found to contain 40% of the carbon-14 that it contained when it was part of a living animal. If the decay of carbon-14 is exponential with an annual rate of decay of 0.0124%, how long ago did the animal die?

24.42

Radioactive strontium-90 is used in nuclear reactors and decays exponentially with an annual rate of decay of 2.48%. How much of 50 grams of strontium-90 will be left after 100 years?

24.43

How long does it take 12 grams of carbon-14 to decay to 10 grams when the decay is exponential with an annual rate of decay of 0.0124%?

24.44

How long does it take for 10 grams of strontium-90 to decay to 8 grams if the decay is exponential and the annual rate of decay is 2.48%?

ANSWERS TO SUPPLEMENTARY PROBLEMS Note: The tables in Appendices A and B were used in computing these answers. If a calculator is used your answers may vary. 24.17

2.7%

24.18

60%

24.19

$6437

24.20

$26 250

24.21

$48 850

24.22

$3820

24.23

$1701

24.24

$2464

24.25

$11 410

24.26

$11 440

24.27

7.68

24.28

6.50

24.29

316 200 000 I0

24.30

177 800 000 I0

CHAP. 24]

APPLICATIONS OF LOGARITHMS AND EXPONENTS

24.31

4.30

24.32

(a) pH ¼ 4:2

24.33

(a) ½Hþ  ¼ 0:001 or 1:00  103

24.34

1.0

24.35

1585 I0

24.36

66 decibels

24.37

6.18 billion

24.38

0.48% per year

24.39

1.31% per year

24.40

3.17% per year

24.41

7390 years

24.42

4.2 grams

24.43

1471 years

24.44

8.998 years

(b) pH ¼ 3:7

(c) pH ¼ 2:2

(d ) pH ¼ 4:1

(b) ½Hþ  ¼ 1:585  108

287

CHAPTER 25

Permutations and Combinations 25.1

FUNDAMENTAL COUNTING PRINCIPLE

If one thing can be done in m different ways and, when it is done in any one of these ways, a second thing can be done in n different ways, then the two things in succession can be done in mn different ways. For example, if there are 3 candidates for governor and 5 for mayor, then the two offices may be filled in 3 . 5 ¼ 15 ways. In general, if a1 can be done in x1 ways, a2 can be done in x2 ways, a3 can be done in x3 ways, . . . , and an can be done in xn ways, then the event a1 a2 a3    an can be done in x1  x2  x3    xn ways. EXAMPLE 25.1. A man has 3 jackets, 10 shirts, and 5 pairs of slacks. If an outfit consists of a jacket, a shirt, and a pair of slacks, how many different outfits can the man make? x1  x2  x3 ¼ 3  10  5 ¼ 150 outfits

25.2

PERMUTATIONS

A permutation is an arrangement of all or part of a number of things in a definite order. For example, the permutations of the three letters a, b, c taken all at a time are abc, acb, bca, bac, cba, cab. The permutations of the three letters a, b, c taken two at a time are ab, ac, ba, bc, ca, cb. For a natural number n, n factorial, denoted by n!, is the product of the first n natural numbers. That is, n! ¼ n  (n  1)  (n  2)    2  1: Also, n! ¼ n  (n  1)! Zero factorial is defined to be 1 : 0! ¼ 1. EXAMPLES 25.2.

Evaluate each factorial.

(a) 7!

(c) 1!

(a) (b) (c) (d ) (e)

(b) 5!

(d ) 2!

(e) 4!

7! ¼ 7 . 6 . 5 . 4 . 3 . 2 . l ¼ 5040 5! ¼ 5 . 4 . 3 . 2 . l ¼ 120 1! ¼ 1 2! ¼ 2 . 1 ¼ 2 4! ¼ 4 . 3 . 2 . 1 ¼ 24

288

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

289

The symbol n Pr represents the number of permutations (arrangements, orders) of n things taken r at a time. Thus 8 P3 denotes the number of permutations of 8 things taken 3 at a time, and 5 P5 denotes the number of permutations of 5 things taken 5 at a time. Note. The symbol P(n, r) having the same meaning as n Pr is sometimes used. A.

Permutations of n different things taken r at a time ¼ n(n  1)(n  2)    (n  r þ 1) ¼

n Pr

n! (n  r)!

When r ¼ n, n Pr ¼ n Pn ¼ n(n  1)(n  2)    1 ¼ n!: EXAMPLES 25.3. 5 P1

¼ 5, 5 P2 ¼ 5  4 ¼ 20, 5 P3 ¼ 5  4  3 ¼ 60, 5 P4 ¼ 5  4  3  2 ¼ 120, 5 P5 ¼ 5! ¼ 5  4  3  2  1 ¼ 120, 10 P7 ¼ 10  9  8  7  6  5  4 ¼ 604 800.

The number of ways in which 4 persons can take their places in a cab having 6 seats is 6 P4 ¼ 6  5  4  3 ¼ 360.

B.

Permutations with some things alike, taken all at a time The number of permutations P of n things taken all at a time, of which n1 are alike, n2 others are alike, n3 others are alike, etc., is P¼

n! n1 ! n2 ! n3 !   

where n1 þ n2 þ n3 þ    ¼ n:

For example, the number of ways 3 dimes and 7 quarters can be distributed among 10 boys, each to receive one coin, is 10! 10  9  8 ¼ ¼ 120: 3! 7! 123 C.

Circular permutations The number of ways of arranging n different objects around a circle is (n  1)! ways. Thus 10 persons may be seated at a round table in (10  1)! ¼ 9! ways.

25.3

COMBINATIONS

A combination is a grouping or selection of all or part of a number of things without reference to the arrangement of the things selected. Thus the combinations of the three letters a, b, c taken 2 at a time are ab, ac, bc. Note that ab and ba are 1 combination but 2 permutations of the letters a, b. The symbol n Cr represents the number of combinations (selections, groups) of n things taken r at a time. Thus 9 C4 denotes the number of combinations of 9 things taken 4 at a time. Note. The symbol C(n, r) having the same meaning as n Cr is sometimes used. A.

Combinations of n different things taken r at a time n Cr

¼

n Pr

r!

¼

n! n(n  1)(n  2)    (n  r þ 1) ¼ r!(n  r)! r!

290

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

For example, the number of handshakes that may be exchanged among a party of 12 students if each student shakes hands once with each other student is 12 C2

¼

12! 12! 12  11 ¼ ¼ ¼ 66: 2!(12  2)! 2!10! 12

The following formula is very useful in simplifying calculations: n Cr

¼ n Cnr :

This formula indicates that the number of selections of r out of n things is the same as the number of selections of n  r out of n things. EXAMPLES 25.4. 5 ¼ 5, 1

5 C1

¼

9 C7

¼ 9 C97

54 ¼ 10, 12 98 ¼ 9 C2 ¼ ¼ 36, 12 5 C2

¼

5 C5

25 C22

¼

5! ¼1 5!

¼ 25 C3 ¼

25  24  23 ¼ 2300 123

Note that in each case the numerator and denominator have the same number of factors.

B.

Combinations of different things taken any number at a time The total number of combinations C of n different things taken 1, 2, 3, . . . , n at a time is C ¼ 2n  1: For example, a woman has in her pocket a quarter, a dime, a nickel, and a penny. The total number of ways she can draw a sum of money from her pocket is 24  1 ¼ 15.

25.4

USING A CALCULATOR

Scientific and graphing calculators have keys for factorials, n!; permutations, n Pr , and combinations, n Cr . As factorials get larger, the results are displayed in scientific notation. Many calculators have only two digits available for the exponent, which limits the size of the factorial that can be displayed. Thus, 69! can be displayed and 70! cannot, because 70! needs more than two digits for the exponent in scientific notation. When the calculator can perform an operation, but it cannot display the result an error message is displayed instead of an answer. The values of n Pr and n Cr can often be computed on the calculator when n! cannot be displayed. This can be done because the internal procedure does not require the result to be displayed, just used.

Solved Problems 25.1

Evaluate

20 P2 , 8 P5 , 7 P5 , 7 P7 .

SOLUTION ¼ 20  19 ¼ 380 P ¼ 8  7  6  5  4 ¼ 6720 8 5

20 P2

25.2

¼ 7  6  5  4  3 ¼ 2520 P ¼ 7! ¼ 7  6  5  4  3  2  1 ¼ 5040 7 7 7 P5

Find n if (a) 7  n P3 ¼ 6  nþ1 P3 ,

(b) 3  n P4 ¼ n1 P5 

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

291

SOLUTION (a) 7n(n  1)(n  2) ¼ 6(n þ 1)(n)(n  1): Since n = 0, 1 we may divide by n(n  1) to obtain 7(n  2) ¼ 6(n þ 1), n ¼ 20. (b) 3n(n  1)(n  2)(n  3) ¼ (n  1)(n  2)(n  3)(n  4)(n  5). Since n=1, 2, 3 we may divide by (n  1)(n  2)(n  3) to obtain 3n ¼ (n  4)(n  5),

n2  12n þ 20 ¼ 0,

(n  10)(n  2) ¼ 0:

Thus n ¼ 10.

25.3

A student has a choice of 5 foreign languages and 4 sciences. In how many ways can he choose 1 language and 1 science? SOLUTION He can choose a language in 5 ways, and with each of these choices there are 4 ways of choosing a science. Hence the required number of ways ¼ 5  4 ¼ 20 ways.

25.4

In how many ways can 2 different prizes be awarded among 10 contestants if both prizes (a) may not be given to the same person, (b) may be given to the same person? SOLUTION (a) The first prize can be awarded in 10 different ways and, when it is awarded, the second prize can be given in 9 ways, since both prizes may not be given to the same contestant. Hence the required number of ways ¼ 10  9 ¼ 90 ways. (b) The first prize can be awarded in 10 ways, and the second prize also in 10 ways, since both prizes may be given to the same contestant. Hence the required number of ways ¼ 10  10 ¼ 100 ways.

25.5

In how many ways can 5 letters be mailed if there are 3 mailboxes available? SOLUTION Each of the 5 letters may be mailed in any of the 3 mailboxes. Hence the required number of ways ¼ 3  3  3  3  3 ¼ 35 ¼ 243 ways.

25.6

There are 4 candidates for president of a club, 6 for vice-president and 2 for secretary. In how many ways can these three positions be filled? SOLUTION A president may be selected in 4 ways, a vice-president in 6 ways, and a secretary in 2 ways. Hence the required number of ways ¼ 4  6  2 ¼ 48 ways.

25.7

In how many different orders may 5 persons be seated in a row? SOLUTION The first person may take any one of 5 seats, and after the first person is seated, the second person may take any one of the remaining 4 seats, etc. Hence the required number of orders ¼ 5  4  3  2  1 ¼ 120 orders. Otherwise.

Number of orders ¼ number of arrangements of 5 persons taken all at a time ¼ 5 P5 ¼ 5! ¼ 5  4  3  2  1 ¼ 120 orders.

292

25.8

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

In how many ways can 7 books be arranged on a shelf? SOLUTION Number of ways ¼ number of arrangements of 7 books taken all at a time ¼ 7 P7 ¼ 7! ¼ 7  6  5  4  3  2  1 ¼ 5040 ways.

25.9

Twelve different pictures are available, of which 4 are to be hung in a row. In how many ways can this be done? SOLUTION The first place may be occupied by any one of 12 pictures, the second place by any one of 11, the third place by any one of 10, and the fourth place by any one of 9. Hence the required number of ways ¼ 12  11  10  9 ¼ 11 880 ways. Otherwise. Number of ways ¼ number of arrangements of 12 pictures taken 4 at a time ¼ 12 P4 ¼ 12  11  10  9 ¼ 11 880 ways:

25.10 It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible? SOLUTION The men may be seated in 5 P5 ways, and the women in 4 P4 ways. Each arrangement of the men may be associated with each arrangement of the women. Hence the required number of arrangements ¼ 5 P5  4 P4 ¼ 5! 4! ¼ 120  24 ¼ 2880:

25.11 In how many orders can 7 different pictures be hung in a row so that 1 specified picture is (a) at the center, (b) at either end? SOLUTION (a) Since 1 given picture is to be at the center, 6 pictures remain to be arranged in a row. Hence the number of orders ¼ 6 P6 ¼ 6! ¼ 720 orders. (b) After the specified picture is hung in any one of 2 ways, the remaining 6 can be arranged in 6 P6 ways. Hence the number of orders ¼ 2  6 P6 ¼ 1440 orders.

25.12 In how many ways can 9 different books be arranged on a shelf so that (a) 3 of the books are always together, (b) 3 of the books are never all 3 together? SOLUTION (a) The specified 3 books can be arranged among themselves in 3 P3 ways. Since the specified 3 books, are always together, they may be considered as 1 thing. Then together with the other 6 books (things) we have a total of 7 things which can be arranged in 7 P7 ways. Total number of ways ¼ 3 P3  7 P7 ¼ 3!7! ¼ 6  5040 ¼ 30 240 ways. (b) Number of ways in which 9 books can be arranged on a shelf if there are no restrictions ¼ 9! ¼ 362 880 ways. Number of ways in which 9 books can be arranged on a shelf when 3 specified books are always together (from (a) above) ¼ 3!7! ¼ 30 240 ways. Hence the number of ways in which 9 books can be arranged on a shelf so that 3 specified books are never all 3 together ¼ 362 880  30 240 ¼ 332 640 ways.

25.13 In how many ways can n women be seated in a row so that 2 particular women will not be next to each other? SOLUTION With no restrictions, n women may be seated in a row in n Pn ways. If 2 of the n women must always sit next to each other, the number of arrangements ¼ 2!(n1 Pn1 ):

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

293

Hence the number of ways n women can be seated in a row if 2 particular women may never sit together ¼ n Pn  2(n1 Pn1 ) ¼ n!  2(n  1)! ¼ n(n  1)!  2(n  1)! ¼ (n  2)  (n  1)!

25.14 Six different biology books, 5 different chemistry books and 2 different physics books are to be arranged on a shelf so that the biology books stand together, the chemistry books stand together, and the physics books stand together. How many such arrangements are possible? SOLUTION The biology books can be arranged among themselves in 6! ways, the chemistry books in 5! ways, the physics books in 2! ways, and the three groups in 3! ways. Required number of arrangements ¼ 6!5!2!3! ¼ 1 036 800.

25.15 Determine the number of different words of 5 letters each that can be formed with the letters of the word chromate (a) if each letter is used not more than once, (b) if each letter may be repeated in any arrangement. (These words need not have meaning.) SOLUTION (a) Number of words ¼ arrangements of 8 different letters taken 5 at a time ¼ 8 P5 ¼ 8  7  6  5  4 ¼ 6720 words. (b) Number of words ¼ 8  8  8  8  8 ¼ 85 ¼ 32 768 words.

25.16 How many numbers may be formed by using 4 out of the 5 digits 1, 2, 3, 4, 5 (a) if the digits must not be repeated in any number, (b) if they may be repeated? If the digits must not be repeated, how many of the 4-digit numbers (c) begin with 2, (d ) end with 25? SOLUTION (a) Numbers formed ¼ 5 P4 ¼ 5  4  3  2 ¼ 120 numbers. (b) Numbers formed ¼ 5  5  5  5 ¼ 54 ¼ 625 numbers. (c) Since the first digit of each number is specified, there remain 4 digits to be arranged in 3 places. Numbers formed ¼ 4 P3 ¼ 4  3  2 ¼ 24 numbers. (d ) Since the last two digits of every number are specified, there remain 3 digits to be arranged in 2 places. Numbers formed ¼ 3 P2 ¼ 3  2 ¼ 6 numbers.

25.17 How many 4-digit numbers may be formed with the 10 digits 0,1, 2, 3, . . . , 9 (a) if each digit is used only once in each number? (b) How many of these numbers are odd? SOLUTION (a) The first place may be filled by any one of the 10 digits except 0, i.e., by any one of 9 digits. The 9 digits remaining may be arranged in the 3 other places in 9 P3 ways. Numbers formed ¼ 9  9 P3 ¼ 9(9  8  7) ¼ 4536 numbers. (b) The last place may be filled by any one of the 5 odd digits, 1,3,5,7,9. The first place may be filled by any one of the 8 digits, i.e., by the remaining 4 odd digits and the even digits, 2,4,6,8. The 8 remaining digits may be arranged in the 2 middle positions in 8 P2 ways. Numbers formed ¼ 5  8  8 P2 ¼ 5  8  8  7 ¼ 2240 odd numbers.

25.18 (a) How many 5-digit numbers can be formed from the 10 digits 0, 1, 2, 3, . . . , 9, repetitions allowed? How many of these numbers (b) begin with 40, (c) are even, (d ) are divisible by 5?

294

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

SOLUTION (a) The first place may be filled by any one of 9 digits (any of the 10 except 0). Each of the other 4 places may be filled by any one of the 10 digits whatever. Numbers formed ¼ 9  10  10  10  10 ¼ 9  104 ¼ 90 000 numbers. (b) The first 2 places may be filled in 1 way, by 40. The other 3 places may be filled by any one of the 10 digits whatever. Numbers formed ¼ 1  10  10  10 ¼ 103 ¼ 1000 numbers. (c) The first place may be filled in 9 ways, and the last place in 5 ways (0, 2, 4, 6, 8). Each of the other 3 places may be filled by any one of the 10 digits whatever. Even numbers ¼ 9  10  10  10  5 ¼ 45 000 numbers. (d ) The first place may be filled in 9 ways, the last place in 2 ways (0, 5), and the other 3 places in 10 ways each. Numbers divisible by 5 ¼ 9  10  10  10  2 ¼ 18 000 numbers.

25.19 How many numbers between 3000 and 5000 can be formed by using the 7 digits 0, 1, 2, 3, 4, 5, 6 if each digit must not be repeated in any number? SOLUTION Since the numbers are between 3000 and 5000, they consist of 4 digits. The first place may be filled in 2 ways, i.e., by digits 3, 4. Then the remaining 6 digits may be arranged in the 3 other places in 6 P3 ways. Numbers formed ¼ 2  6 P3 ¼ 2(6  5  4) ¼ 240 numbers.

25.20 From 11 novels and 3 dictionaries, 4 novels and 1 dictionary are to be selected and arranged on a shelf so that the dictionary is always in the middle. How many such arrangements are possible? SOLUTION The dictionary may be chosen in 3 ways. The number of arrangements of 11 novels taken 4 at a time is Required number of arrangements ¼ 3  11 P4 ¼ 3(11  10  9  8) ¼ 23 760.

11 P4 .

25.21 How many signals can be made with 5 different flags by raising them any number at a time? SOLUTION Signals may be made by raising the flags 1, 2, 3, 4, and 5 at a time. Hence the total number of signals is 5 P1

þ 5 P2 þ 5 P3 þ 5 P4 þ 5 P5 ¼ 5 þ 20 þ 60 þ 120 þ 120 ¼ 325 signals:

25.22 Compute the sum of the 4-digit numbers which can be formed with the four digits 2, 5, 3, 8 if each digit is used only once in each arrangement. SOLUTION The number of arrangements, or numbers, is 4 P4 ¼ 4! ¼ 4  3  2  1 ¼ 24: The sum of the digits ¼ 2 þ 5 þ 3 þ 8 ¼ 18, and each digit will occur 24=4 ¼ 6 times each in the units, tens, hundreds, and thousands positions. Hence the sum of all the numbers formed is 1(6  18) þ 10(6  18) þ 100(6  18) þ 1000(6  18) ¼ 119 988:

25.23 (a) How many arrangements can be made from the letters of the word cooperator when all are taken at a time? How many of such arrangements (b) have the three os together, (c) begin with the two rs?

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

295

SOLUTION (a) The word cooperator consists of 10 letters: 3 os, 2 rs, and 5 different letters. Number of arrangements ¼

10! 10  9  8  7  6  5  4  3  2  1 ¼ ¼ 302 400: 3!2! (1  2  3)(1  2)

(b) Consider the 3 os as 1 letter. Then we have 8 letters of which 2 rs are alike. Number of arrangements ¼

8! ¼ 20 160: 2!

(c) The number of arrangements of the remaining 8 letters, of which 3 os are alike, ¼ 8!=3! ¼ 6720.

25.24 There are 3 copies each of 4 different books. In how many different ways can they be arranged on a shelf? SOLUTION There are 3  4 ¼ 12 books of which 3 are alike, 3 others alike, etc. Number of arrangements ¼

(3  4)! 12! ¼ ¼ 369 600: 3!3!3!3! (3!Þ4

25.25 (a) In how many ways can 5 persons be seated at a round table? (b) In how many ways can 8 persons be seated at a round table if 2 particular persons must always sit together? SOLUTION (a) Let 1 of them be seated anywhere. Then the 4 persons remaining can be seated in 4! ways. Hence there are 4! ¼ 24 ways of arranging 5 persons in a circle. (b) Consider the two particular persons as one person. Since there are 2! ways of arranging 2 persons among themselves and 6! ways of arranging 7 persons in a circle, the required number of ways ¼ 2!6! ¼ 2  720 ¼ 1440 ways.

25.26 In how many ways can 4 men and 4 women be seated at a round table if each woman is to be between two men? SOLUTION Consider that the men are seated first. Then the men can be arranged in 3! ways, and the women in 4! ways. Required number of circular arrangements ¼ 3!4! ¼ 144.

25.27 By stringing together 9 differently colored beads, how many different bracelets can be made? SOLUTION There are 8! arrangements of the beads on the bracelet, but half of these can be obtained from the other half simply by turning the bracelet over. Hence there are 12 (8!) ¼ 20 160 different bracelets.

25.28 In each case, find n: (a) n Cn2 ¼ 10, (b) n C15 ¼ n C11 , (c) n P4 ¼ 30  nC5 . SOLUTION (a)

n Cn2

¼ n C2 ¼

n(n  1) n2  n ¼ ¼ 10, 2! 2

n2  n  20 ¼ 0,

n¼5

296

PERMUTATIONS AND COMBINATIONS

(b)

n Cr

¼ n Cnr ,

(c) 30  n C5 ¼ 30 Then

n P4

¼ n Cn11 ,

n C15



n P5

5! ¼

 ¼

15 ¼ n  11,

[CHAP. 25

n ¼ 26

30  n P4  (n  4) 5!

30  n P4  (n  4) , 5!

30(n  4) , 120

1¼

n ¼ 8:

25.29 Given n Pr ¼ 3024 and n Cr ¼ 126, find r. SOLUTION n Pr

¼ r!(n Cr ),

r! ¼

n Pr n Cr

¼

3024 ¼ 24, 126

r¼4

25.30 How many different sets of 4 students can be chosen out of 17 qualified students to represent a school in a mathematics contest? SOLUTION Number of sets ¼ number of combinations of 4 out of 17 students ¼ 17 C4 ¼

17  16  15  14 ¼ 2380 sets of 4 students. 1234

25.31 In how many ways can 5 styles be selected out of 8 styles? SOLUTION Number of ways ¼ number of combinations of 5 out of 8 styles ¼ 8 C5 ¼ 8 C3 ¼

876 ¼ 56 ways: 123

25.32 In how many ways can 12 books be divided between A and B so that one may get 9 and the other 3 books? SOLUTION In each separation of 12 books into 9 and 3, A may get the 9 and B the 3, or A may get the 3 and B the 9.   12  11  10 Hence the number of ways ¼ 2  12 C9 ¼ 2  12 C3 ¼ 2 ¼ 440 ways. 123

25.33 Determine the number of different triangles which can be formed by joining the six vertices of a hexagon, the vertices of each triangle being on the hexagon. SOLUTION Number of triangles ¼ number of combinations of 3 out of 6 points ¼ 6 C3 ¼

654 ¼ 20 triangles: 123

25.34 How many angles less than 1808 are formed by 12 straight lines which terminate in a point, if no two of them are in the same straight line?

CHAP. 25]

297

PERMUTATIONS AND COMBINATIONS

SOLUTION Number of angles ¼ number of combinations of 2 out of 12 lines ¼ 12 C2 ¼

12  11 ¼ 66 angles. 12

25.35 How many diagonals has an octagon? SOLUTION

87 ¼ 28. 2 Since 8 of these 28 lines are the sides of the octagon, the number of diagonals ¼ 20. Lines formed ¼ number of combinations of 2 out of 8 corners (points) ¼ 8 C2 ¼

25.36 How many parallelograms are formed by a set of 4 parallel lines intersecting another set of 7 parallel lines? SOLUTION Each combination of 2 lines out of 4 can intersect each combination of 2 lines out of 7 to form a parallelogram. Number of parallelograms ¼ 4 C2  7 C2 ¼ 6  21 ¼ 126 parallelograms.

25.37 There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points? SOLUTION Number of lines formed if no 3 of the 10 points were in a straight line ¼ 10 C2 ¼ Number of lines formed by 4 points, no 3 of which are collinear ¼ 4 C2 ¼

10  9 ¼ 45. 2

43 ¼ 6. 2

Since the 4 points are collinear, they form 1 line instead of 6 lines. Required number of lines ¼ 45  6 þ 1 ¼ 40 lines.

25.38 In how many ways can 3 women be selected out of 15 women? (a) if 1 of the women is to be included in every selection, (b) if 2 of the women are to be excluded from every selection, (c) if 1 is always included and 2 are always excluded? SOLUTION (a) Since 1 is always included, we must select 2 out of 14 women. Hence the number of ways ¼ 14 C2 ¼

14  13 ¼ 91 ways: 2

(b) Since 2 are always excluded, we must select 3 out of 13 women. Hence the number of ways ¼ 13 C3 ¼ (c) Number of ways ¼ 1512 C31 ¼ 12 C2 ¼

13  12  11 ¼ 286 ways: 3!

12  11 ¼ 66 ways. 2

25.39 An organization has 25 members, 4 of whom are doctors. In how many ways can a committee of 3 members be selected so as to include at least 1 doctor?

298

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

SOLUTION Total number of ways in which 3 can be selected out of 25 ¼ 25 C3 : Number of ways in which 3 can be selected so that no doctor is included ¼ 254 C3 ¼ 21 C3 . Then the number of ways in which 3 members can be selected so that at least 1 doctor is included is 25 C3

 21 C3 ¼

25  24  23 21  20  19  ¼ 970 ways: 3! 3!

25.40 From 6 chemists and 5 biologists, a committee of 7 is to be chosen so as to include 4 chemists. In how many ways can this be done? SOLUTION Each selection of 4 out of 6 chemists can be associated with each selection of 3 out of 5 biologists. Hence the number of ways ¼ 6 C4  5 C3 ¼ 6 C2  5 C2 ¼ 15  10 ¼ 150 ways.

25.41 Given 8 consonants and 4 vowels, how many 5-letter words can be formed, each word consisting of 3 different consonants and 2 different vowels? SOLUTION The 3 different consonants can be selected in 8 C3 ways, the 2 different vowels in 4 C2 ways, and the 5 different letters (3 consonants, 2 vowels) can be arranged among themselves in 5 P5 ¼ 5! ways. Hence the number of words ¼ 8 C3  4 C2  5! ¼ 56  6  120 ¼ 40 320.

25.42 From 7 capitals, 3 vowels and 5 consonants, how many words of 4 letters each can be formed if each word begins with a capital and contains at least 1 vowel, all the letters of each word being different? SOLUTION The first letter, or capital, may be selected in 7 ways. The remaining 3 letters may be (a) 1 vowel and 2 consonants, which may be selected in 3 C1  5 C2 ways, (b) 2 vowels and 1 consonant, which may be selected in 3 C2  5 C1 ways, and (c) 3 vowels, which may be selected in 3 C3 ¼ 1 way. Each of these selections of 3 letters may be arranged among themselves in 3 P3 ¼ 3! ways. Hence the number of words ¼ 7  3!(3 C1  5 C2 þ 3 C2  5 C1 þ 1) ¼ 7  6(3  10 þ 3  5 þ 1) ¼ 1932 words:

25.43 A has 3 maps and B has 9 maps. Determine the number of ways in which they can exchange maps if each keeps his initial number of maps. SOLUTION A can exchange 1 map with B in 3 C1  9 C1 ¼ 3  9 ¼ 27 ways. A can exchange 2 maps with B in 3 C2  9 C2 ¼ 3  36 ¼ 108 ways. A can exchange 3 maps with B in 3 C3  9 C3 ¼ 1  84 ¼ 84 ways. Total number of ways ¼ 27 þ 108 þ 84 ¼ 219 ways. Another method. Consider that A and B put their maps together. Then the problem is to find the number of ways A can select 3 maps out of 12, not including the selection by A of his original three maps. Hence,

12 C3

1¼

12  11  10  1 ¼ 219 ways: 123

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

299

25.44 (a) In how many ways can 12 books be divided among 3 students so that each receives 4 books? (b) In how many ways can 12 books be divided into 3 groups of 4 each? SOLUTION (a) The first student can select 4 out of 12 books in 12 C4 ways. The second student can select 4 of the remaining 8 books in 8 C4 ways. The third student can select 4 of the remaining 4 books in 1 way. Number of ways ¼

12 C4

 8 C4  1 ¼ 495  70  1 ¼ 34 650 ways:

(b) The 3 groups could be distributed among the students in 3! ¼ 6 ways. Hence the number of groups ¼ 34 650=3! ¼ 5775 groups:

25.45 In how many ways can a person choose 1 or more of 4 electrical appliances? SOLUTION Each appliance may be dealt with in 2 ways, as it can be chosen or not chosen. Since each of the 2 ways of dealing with an appliance is associated with 2 ways of dealing with each of the other appliances, the number of ways of dealing with the 4 appliances ¼ 2  2  2  2 ¼ 24 ways. But 24 ways includes the case in which no appliance is chosen. Hence the required number of ways ¼ 24  1 ¼ 16  1 ¼ 15 ways. Another method. The appliances may be chosen singly, in twos, etc. Hence the required number of ways ¼ 4 C1 þ 4 C2 þ 4 C3 þ 4 C4 ¼ 4 þ 6 þ 4 þ 1 ¼ 15 ways.

25.46 How many different sums of money can be drawn from a wallet containing one bill each of 1, 2, 5, 10, 20 and 50 dollars? SOLUTION Number of sums ¼ 26  1 ¼ 63 sums:

25.47 In how many ways can 2 or more ties be selected out of 8 ties? SOLUTION One or more ties may be selected in (28  1) ways. But since 2 or more must be chosen, the required number of ways ¼ 28  1  8 ¼ 247 ways. Another method. 2, 3, 4, 5, 6, 7, or 8 ties may be selected in 8 C2

þ 8C3 þ 8C4 þ 8C5 þ 8C6 þ 8C7 þ 8C8 ¼ 8 C2 þ 8C3 þ 8C4 þ 8C3 þ 8C2 þ 8C1 þ 1 ¼ 28 þ 56 þ 70 þ 56 þ 28 þ 8 þ 1 ¼ 247 ways:

25.48 There are available 5 different green dyes, 4 different blue dyes, and 3 different red dyes. How many selections of dyes can be made, taking at least 1 green and 1 blue dye? SOLUTION The green dyes can be chosen in (25  1) ways, the blue dyes in (24  1) ways, and the red dyes in 23 ways. Number of selections ¼ (25  1)(24  1)(23 ) ¼ 31  15  8 ¼ 3720 selections:

300

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

Supplementary Problems 16 P3 , 7 P4 , 5 P5 , 12 P1 :

25.49

Evaluate:

25.50

Find n if (a) 10  n P2 ¼ nþ1 P4 , (b) 3  2nþ4 P3 ¼ 2  nþ4 P4 .

25.51

In how many ways can six people be seated on a bench?

25.52

With four signal flags of different colors, how many different signals can be made by displaying two flags one above the other?

25.53

With six signal flags of different colors, how many different signals can be made by displaying three flags one above the other?

25.54

In how many ways can a club consisting of 12 members choose a president, a secretary, and a treasurer?

25.55

If no two books are alike, in how many ways can 2 red, 3 green, and 4 blue books be arranged on a shelf so that all the books of the same color are together?

25.56

There are 4 hooks on a wall. In how many ways can 3 coats be hung on them, one coat on a hook?

25.57

How many two-digit numbers can be formed with the digits 0, 3, 5, 7 if no repetition in any of the numbers is allowed?

25.58

How many even numbers of two different digits can be formed from the digits 3, 4, 5, 6, 8?

25.59

How many three-digit numbers can be formed from the digits 1, 2, 3, 4, 5 if no digit is repeated in any number?

25.60

How many numbers of three digits each can be written with the digits 1, 2, . . . , 9 if no digit is repeated in any number?

25.61

How many three-digit numbers can be formed from the digits 3, 4, 5, 6, 7 if digits are allowed to be repeated?

25.62

How many odd numbers of three digits each can be formed, without the repetition of any digit in a number, from the digits (a) 1, 2, 3, 4, (b) 1, 2, 4, 6, 8?

25.63

How many even numbers of four different digits each can be formed from the digits 3, 5, 6, 7, 9?

25.64

How many different numbers of 5 digits each can be formed from the digits 2, 3, 5, 7, 9 if no digit is repeated?

25.65

How many integers are there between 100 and 1000 in which no digit is repeated?

25.66

How many integers greater than 300 and less than 1000 can be made with the digits 1, 2, 3, 4, 5 if no digit is repeated in any number?

25.67

How many numbers between 100 and 1000 can be written with the digits 0, 1, 2, 3, 4 if no digit is repeated in any number?

25.68

How many four-digit numbers greater than 2000 can be formed with the digits 1, 2, 3, 4 if repetitions (a) are not allowed, (b) are allowed?

25.69

How many of the arrangements of the letters of the word logarithm begin with a vowel and end with a consonant?

25.70

In a telephone system four different letters P, R, S, T and the four digits 3, 5, 7, 8 are used. Find the maximum number of “telephone numbers” the system can have if each consists of a letter followed by a four-digit number in which the digits may be repeated.

CHAP. 25]

PERMUTATIONS AND COMBINATIONS

301

25.71

In how many ways can 3 girls and 3 boys be seated in a row, if no two girls and no two boys are to occupy adjacent seats?

25.72

How many four character codes can be made by using three dots and two dashes?

25.73

In how many ways can three dice fall?

25.74

How many fraternities can be named with the 24 letters of the Greek alphabet if each has three letters and none is repeated in any name?

25.75

How many signals can be shown with 8 flags of which 2 are red, 3 white and 3 blue, if they are all strung up on a vertical pole at once?

25.76

In how many ways can 4 men and 4 women sit at a round table so that no two men are adjacent?

25.77

How many different arrangements are possible with the factors of the term a2 b4 c5 written at full length?

25.78

In how many ways can 9 different prizes be awarded to two students so that one receives 3 and the other 6?

25.79

How many different radio stations can be named with 3 different letters of the alphabet? How many with 4 different letters in which W must come first?

25.80

In each case find n: (a) 4  n C2 ¼

25.81

If 5  n P3 ¼ 24  n C4 , find n.

25.82

Evaluate (a) 7 C7 , (b) 5 C3 , (c) 7 C2 , (d ) 7 C5 , (e) 7 C6 , ( f ) 8 C7 , (g) 8 C5 , (h)

25.83

How many straight lines are determined by (a) 6, (b) n points, no three of which lie in the same straight line?

25.84

How many chords are determined by seven points on a circle?

25.85

A student is allowed to choose 5 questions out of 9. In how many ways can she choose them?

25.86

How many different sums of money can be formed by taking two of the following: a cent, a nickel, a dime, a quarter, a half-dollar?

25.87

How many different sums of money can be formed from the coins of Problem 25.86?

25.88

A baseball league is made up of 6 teams. If each team is to play each of the other teams (a) twice, (b) three times, how many games will be played?

25.89

How many different committees of two men and one woman can be formed from (a) 7 men and 4 women, (b) 5 men and 3 women?

25.90

In how many ways can 5 colors be selected out of 8 different colors including red, blue, and green (a) if blue and green are always to be included, (b) if red is always excluded, (c) if red and blue are always included but green excluded?

25.91

From 5 physicists, 4 chemists, and 3 mathematicians a committee of 6 is to be chosen so as to include 3 physicists, 2 chemists, and 1 mathematician. In how many ways can this be done?

25.92

In Problem 25.91, in how many ways can the committee of 6 be chosen so that

nþ2 C3 ,

(b)

nþ2 Cn

¼ 45,

(c) nC12 ¼ n C8 .

100 C98 .

(a) 2 members of the committee are mathematicians. (b) at least 3 members of the committee are physicists? 25.93

How many words of 2 vowels and 3 consonants may be formed (considering any set a word) from the letters of the word (a) stenographic, (b) facetious?

302

PERMUTATIONS AND COMBINATIONS

[CHAP. 25

25.94

In how many ways can a picture be colored if 7 different colors are available for use?

25.95

In how many ways can 8 women form a committee if at least 3 women are to be on the committee?

25.96

A box contains 7 red cards, 6 white cards and 4 blue cards. How many selections of three cards can be made so that (a) all three are red, (b) none is red?

25.97

How many teams of nine players can be chosen from 13 candidates if A, B, C, D are the only candidates for two positions and can play no other position?

25.98

How many different committees including 3 Democrats and 2 Republicans can be chosen from 8 Republicans and 10 Democrats?

25.99

At a meeting, after everyone had shaken hands once with everyone else, it was found that 45 handshakes were exchanged. How many were at the meeting?

ANSWERS TO SUPPLEMENTARY PROBLEMS 25.83 (a) 15, (b)

n(n  1) 2

25.49

3360, 840, 120, 12

25.67 48

25.50

(a) 4, (b) 6

25.68 (a) 18,

25.51

720

25.69 90 720

25.85 126

25.52

12

25.70 1024

25.86 10

25.53

120

25.71 72

25.87 31

25.54

1320

25.72 10

25.88 (a) 30, (b) 45

25.55

1728

25.73 216

25.89 (a) 84, (b) 30

25.56

24

25.74 12, 144

25.90 (a) 20, (b) 21, (c) 10

25.57

9

25.75 560

25.91 180

25.58

12

25.76 144

25.92 (a) 378, (b) 462

25.59

60

25.77 6930

25.93 (a) 40 320, (b) 4800

25.60

504

25.78 168

25.94 127

25.61

125

25.79 15 600; 13 800

25.95 219

25.62

(a) 12, (b) 12

25.80 (a) 2, 7, (b) 8, (c) 20

25.96 (a) 35, (b) 120

25.63

24

25.81 8

25.97 216

25.64

120

25.82 (a) 1, (b) 10, (c) 21,

25.98 3360

25.65

648

(d ) 21, (e) 7, ( f ) 8,

25.66

36

(g) 56, (h) 4950

(b) 192

25.84 21

25.99 10

CHAPTER 26

The Binomial Theorem 26.1

COMBINATORIAL NOTATION

The number of combinations of n objects selected r at a time, n Cr , can be written in the form   n r which is called combinatorial notation. n! ¼ n Cr ¼ (n  r)!r!

  n ; r

where n and r are integers and r  n. EXAMPLES 26.1. Evaluate each expression.         7 8 9 5 (a) (b) (c) (d ) 3 7 9 0   7! 7! 7  6  5  4! 7 ¼ ¼ ¼ 7  5 ¼ 35 ¼ 3 (7  3)!3! 4!3! 4!3  2  1   8! 8! 8  7! 8 ¼ ¼ ¼8 (b) ¼ 7 (8  7)!7! 1!7! 1  17!   9! 9! 1 1 9 ¼ ¼ ¼ ¼1 (c) ¼ 9 (9  9)!9! 0!9! 0! 1   5! 5! 1 1 5 ¼ ¼ ¼ ¼1 (d ) ¼ 0 (5  0)!0! 5!0! 0! 1 (a)

26.2

EXPANSION OF (a þ x)n

If n is a positive integer we expand (a þ x)n as shown below: n(n  1) n2 2 n(n  1)(n  2) n3 3 a x þ a x 2! 3! n(n  1)(n  2)    (n  r þ 2) nrþ1 r1 a þ  þ x þ    þ xn (r  1)! This equation is called the binomial theorem, or binomial formula. (a þ x)n ¼ an þ nan1 x þ

303

304

THE BINOMIAL THEOREM

[CHAP. 26

Other forms of the binomial theorem exist and some use combinations to express the coefficients. The relationship between the coefficients and combinations are shown below.   5 54 54321 5! 5! ¼ ¼ ¼ ¼ 2! 3  2  1  2! 3!2! (5  2)!2! 2   n n(n  1)(n  2) n(n  1)(n  2)    2  1 n! ¼ ¼ ¼ 3! (n  3)!3! (n  3)!3! 3 So (a þ x)n ¼ an þ

n! n! an1 x þ an2 x2 þ    (n  1)!1! (n  2)!2!

n! anrþ1 xr1 þ    þ xn (n  ½r  1)!(r  1)!       n n1 n n2 2 n n n (a þ x) ¼ a þ a xþ a x þ  þ anrþ1 xr1 þ    þ xn 1 2 r1 þ

and

The rth term of the expansion of (a þ x)n is rth term ¼

n(n  1)(n  2)    (n  r þ 2) nrþ1 r1 a x : (r  1)!

The rth term formula for the expansion of (a þ x)n can be expressed in terms of combinations. rth term ¼ ¼

n(n  1)(n  2)    (n  r þ 2) nrþ1 r1 a x (r  1)! n(n  1)(n  2)    (n  r þ 2)(n  r þ 1)    2  1 nrþ1 r1 a x (n  r þ 1)(n  r)    2  1(r  1)!

n! anrþ1 xr1 (n  ½r  1)!(r  1)!   n rth term ¼ anrþ1 xr1 r1 rth term ¼

Solved Problems 26.1

Evaluate each expression.     10 10 (a) (b) 8 2

 (c)

12 10



 (d )

SOLUTION   10! 10! 10  9  8! 10 (a) ¼ ¼ ¼ ¼ 45 2 (10  2)!2! 8!2! 8!  2  1   10! 10! 10  9  8! 10 ¼ ¼ ¼ 45 (b) ¼ 8 (10  8)!8! 2!8! 2  1  8!   12! 12! 12  11  10! 12 ¼ ¼ ¼ 66 (c) ¼ 10 (12  10)!10! 2!  10! 2  1  10!   170 170! 1 1 170 ¼ ¼ ¼ ¼1 (d ) ¼ 170 (170  170)!170! 0!  170! 0! 1

170 170



CHAP. 26]

THE BINOMIAL THEOREM

305

Expand by the binomial formula. 26.2

(a þ x)3 ¼ a3 þ 3a2 x þ

32 2 321 3 ax þ x ¼ a3 þ 3a2 x þ 3ax2 þ x3 12 123

26.3

(a þ x)4 ¼ a4 þ 4a3 x þ

43 2 2 432 3 4321 4 ax þ ax þ x ¼ a4 þ 4a3 x þ 6a2 x2 þ 4ax3 þ x4 12 123 1234

26.4

(a þ x)5 ¼ a5 þ 5a4 x þ

54 3 2 543 2 3 5432 2 4 ax þ a x þ a x þ x5 ¼ a5 þ 5a4 x þ 10a3 x2 12 123 1234 þ 10a2 x3 þ 5ax4 þ x5

Note that in the expansion of (a þ x)n : (1) The exponent of a þ the exponent of x ¼ n (i.e., the degree of each term is n). (2) The number of terms is n þ 1, when n is a positive integer. (3) There are two middle terms when n is an odd positive integer. (4) There is only one middle term when n is an even positive integer. (5) The coefficients of the terms which are equidistant from the ends are the same. It is interesting to note that these coefficients may be arranged as follows. (a þ x)0 (a þ x)1 (a þ x)2 (a þ x)3 (a þ x)4 (a þ x)5 (a þ x)6 etc:

1 1 1 1

1 2

3

1 3

1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

This array of numbers is known as Pascal’s Triangle. The first and last number n in each row are 1, while any other number in the array can be obtained by adding the two numbers to the right and left of it in the preceding row. 26.5

(x  y2 )6 ¼ x6 þ 6x5 (y2 ) þ þ

65 4 654 3 6543 2 x (y2 )2 þ x (y2 )3 þ x (y2 )4 12 1  2  3 1234

65432 x(y2 )5 þ (y2 )6 12345

¼ x6  6x5 y2 þ 15x4 y4  20x3 y6 þ 15x2 y8  6xy10 þ y12 In the expansion of a binomial of the form (a  b)n , where n is a positive integer, the terms are alternately þ and 2. 26.6

(3a3  2b)4 ¼ (3a3 )4 þ 4(3a3 )3 (2b) þ

43 432 (3a3 )2 (2b)2 þ (3a3 )(2b)3 þ (2b)4 12 123

¼ 81a12  216a9 b þ 216a6 b2  96a3 b3 þ 16b4 26.7

(x  1)7 ¼ x7 þ 7x6 (1) þ þ

76 5 765 4 7654 3 x (1)2 þ x (1)3 þ x (1)4 12 123 1234

76543 2 765432 x (1)5 þ x(1)6 þ (1)7 12345 123456

¼ x7  7x6 þ 21x5  35x4 þ 35x3  21x2 þ 7x  1

306

THE BINOMIAL THEOREM

 26.8

26.9

x 2 þ 3 y

¼

 4  3         3  4 x x 2 43 x 2 2 2 432 x 2 2 þ4 þ þ þ 3 3 y 12 3 y 123 3 y y

¼

x4 8x3 8x2 32x 16 þ þ þ þ 81 27y 3y2 3y3 y4

4

pffiffiffi pffiffiffi 6  5 1=2 4 1=2 2 6  5  4 1=2 3 1=2 3 (x ) (x ) (x ) ( y ) ( x þ y)6 ¼ (x1=2 )6 þ 6(x1=2 )5 ( y1=2 ) þ 12 123 þ

6  5  4  3 1=2 2 1=2 4 6  5  4  3  2 1=2 1=2 5 (x ) ( y ) þ (x )( y ) þ ( y1=2 )6 1234 12345

¼ x3 þ 6x5=2 y1=2 þ 15x2 y þ 20x3=2 y3=2 þ 15xy2 þ 6x1=2 y5=2 þ y3 26.10 (a2 þ b3=2 )4 ¼ (a2 )4 þ 4(a2 )3 (b3=2 ) þ

4  3 2 2 3=2 2 4  3  2 2 3=2 3 (a ) (b ) þ (a )(b ) þ (b3=2 )4 12 123

¼ a8 þ 4a6 b3=2 þ 6a4 b3 þ 4a2 b9=2 þ b6 26.11 (ex  ex )7 ¼ (ex )7 þ 7(ex )6 (ex ) þ

76 x 5 765 x 4 (e ) (ex )2 þ (e ) (ex )3 12 123

þ

7654 x 3 76543 x 2 (e ) (ex )4 þ (e ) (ex )5 1234 12345

þ

765432 x (e )(ex )6 þ (ex )7 123456

¼ e7x  7e5x þ 21e3x  35ex þ 35ex  21e3x þ 7e5x  e7x 26.12 (a þ b  c)3 ¼ ½(a þ b)  c3 ¼ (a þ b)3 þ 3(a þ b)2 (c) þ

32 (a þ b)(c)2 þ (c)3 12

¼ a3 þ 3a2 b þ 3ab2 þ b3  3a2 c  6abc  3b2 c þ 3ac2 þ 3bc2  c3 26.13 (x2 þ x  3)3 ¼ ½x2 þ (x  3)3 ¼ (x2 )3 þ 3(x2 )2 (x  3) þ

32 2 (x )(x  3)2 þ (x  3)3 12

¼ x6 þ (3x5  9x4 ) þ (3x4  18x3 þ 27x2 ) þ (x3  9x2 þ 27x  27) ¼ x6 þ 3x5  6x4  17x3 þ 18x2 þ 27x  27 In Problems 26.14 – 26.18, write the indicated term of each expansion, using the formula rth term of (a þ x)n ¼

n(n  1)(n  2)    (n  r þ 2) nrþ1 r1 a x : (r  1)!

26.14 Sixth term of (x þ y)15 . SOLUTION n ¼ 15, r ¼ 6, n 2 r þ 2 ¼ 11, r 2 l ¼ 5, n 2 r þ 1 ¼ 10 15  14  13  12  11 10 5 6th term ¼ x y ¼ 3003x10 y5 12345

26.15 Fifth term of (a 

pffiffiffi 9 b) .

[CHAP. 26

CHAP. 26]

THE BINOMIAL THEOREM

SOLUTION n ¼ 9, r ¼ 5, n 2 r þ 2 ¼ 6, r 2 l ¼ 4, n 2 r þ l ¼ 5 9  8  7  6 5 pffiffiffi 4 5th term ¼ a ( b) ¼ 126a5 b2 1234

26.16 Fourth term of (x2  y2 )11 : SOLUTION n ¼ 11, r ¼ 4, n 2 r þ 2 ¼ 9, r 2 1 ¼ 3, n 2 r þ l ¼ 8 4th term ¼



x 1 þ 26.17 Ninth term of 2 x

11  10  9 2 8 (x ) (y2 )3 ¼ 165x16 y6 123

12 :

SOLUTION n ¼ 12, r ¼ 9, n  r þ 2 ¼ 5, r  1 ¼ 8, n  r þ 1 ¼ 4     12  11  10  9  8  7  6  5 x 4 1 8 495 ¼ 9th term ¼ 12345678 2 x 16x4



1 26.18 Eighteenth term of 1  x

20 :

SOLUTION n ¼ 20, r ¼ 18, n  r þ 2 ¼ 4, r  1 ¼ 17, n  r þ 1 ¼ 3  17 20  19  18  17    4 1 20  19  18 1140 18th term = ¼ ¼  17  17 1  2  3  4    17 x 1  2  3x x

26.19 Find the term involving x2 in the expansion of  a10 x3 þ : x SOLUTION From (x3 )10rþ1 (x1 )r1 ¼ x2 we obtain 3(10  r þ 1)  1(r  1) ¼ 2 or r ¼ 8. For the 8th term: n ¼ 10, r ¼ 8, n  r þ 2 ¼ 4, r  1 ¼ 7, n  r þ 1 ¼ 3. 8th term ¼

  10  9  8  7  6  5  4 3 3 a 7 ¼ 120a7 x2 (x ) 1234567 x

26.20 Find the term independent of x in the expansion of   1 9 2 : x  x SOLUTION From (x2 )9rþ1 (x1 )r1 ¼ x0 we obtain 2(9  r þ 1)  1(r  1) ¼ 0 or r ¼ 7.

307

308

THE BINOMIAL THEOREM

[CHAP. 26

For the 7th term: n ¼ 9, r ¼ 7, n  r þ 2 ¼ 4, r  1 ¼ 6, n  r þ 1 ¼ 3. 7th term ¼

987654 2 3 (x ) (x1 )6 ¼ 84 123456

26.21 Evaluate (1:03)10 to five significant figures. SOLUTION (1:03)10 ¼ (1 þ 0:03)10 ¼ 1 þ 10(0:03) þ

10  9 10  9  8 10  9  8  7 (0:03)2 þ (0:03)3 þ (0:03)4 þ    12 123 1234

¼ 1 þ 0:3 þ 0:0405 þ 0:003 24 þ 0:000 17 þ    ¼ 1:3439 Note that all 11 terms of the expansion of (0:03 þ 1)10 would be required in order to evaluate (1:03)10 .

26.22 Evaluate (0:99)15 to four decimal places. SOLUTION (0:99)15 ¼ (1  0:01)15 ¼ 1 þ 15(0:01) þ þ

15  14 15  14  13 (0:01)2 þ (0:01)3 12 123

15  14  13  12 þ (0:01)4 þ    1234

¼ 1  0:15 þ 0:0105  0:000 455 þ 0:000 014     ¼ 0:8601

26.23 Find the sum of the coefficients in the expansion of (a) (1 þ x)10 , (b) (1  x)10 : SOLUTION (a) If, 1, c1 , c2 , . . . , c10 are the coefficients, we have the identity (1 þ x)10 ¼ 1 þ c1 x þ c2 x2 þ    þ c10 x10 :

Let x ¼ 1:

Then (1 þ 1)10 ¼ 1 þ c1 þ c2 þ    þ c10 ¼ sum of coefficients ¼ 210 ¼ 1024 (b) Let x ¼ l. Then (1  x)10 ¼ (1  1)10 ¼ 0 ¼ sum of coefficients.

Supplementary Problems 26.24

Expand by the binomial formula. (a) (x þ 12 )6 (b) (x  2)5

26.25

(c) (y þ 3)4   1 5 (d ) x þ x



 x 3 4 þ 2 y

(e) (x2  y3 )4

(g)

( f ) (a  2b)6

(h) ( y1=2 þ y1=2 )6

Write the indicated term in the expansion of each of the following.   1 10 (a) Fifth term of (a  b)7 (d ) Seventh term of a  pffiffiffi a  9 1 (b) Seventh term of x2  (e) Sixteenth term of (2  1=x)18 x   1 8 (c) Middle term of y  ( f ) Sixth term of (x2  2y)11 y

CHAP. 26]

26.26

THE BINOMIAL THEOREM

Find the term independent of x in the expansion of 

26.27

pffiffiffi 1 x 2 3x

10 :

Find the term involving x3 in the expansion of   1 12 : x2 þ x

26.28

Evaluate (0:98)6 correct to five decimal places.

26.29

Evaluate (1:1)10 correct to the nearest hundredth.

ANSWERS TO SUPPLEMENTARY PROBLEMS 15 5 15 3 1 26.24 (a) x6 þ 3x5 þ x4 þ x3 þ x2 þ x þ 4 2 16 16 64 (b) x5  10x4 þ 40x3  80x2 þ 80x  32 (c) y4 þ 12y3 þ 54y2 þ 108y þ 81 (d ) x5 þ 5x3 þ 10x þ

10 5 1 þ þ x x3 x5

(e) x8  4x6 y3 þ 6x4 y6  4x2 y9 þ y12 ( f ) a6  12a5 b þ 60a4 b2  160a3 b3 þ 240a2 b4  192ab5 þ 64b6 (g)

x4 3x3 27x2 54x 81 þ þ 2 þ 3 þ 4 16 2y 2y y y

(h) y3 þ 6y2 þ 15y þ 20 þ 15y1 þ 6y2 þ y3 26.25

(a) 35a3 b4

(c)

(b) 84

(d ) 210a

26.26

5

26.27

792x3

26.28

0.885 84

26.29

2.59

70

(e) 

6528 x15

( f ) 14 784x12 y5

309

CHAPTER 27

Probability 27.1

SIMPLE PROBABILITY

Suppose that an event can happen in h ways and fail to happen in f ways, all these h þ f ways supposed equally likely. Then the probability of the occurrence of the event (called its success) is p¼

h h ¼ , hþf n

and the probability of the non-occurrence of the event (called its failure) is q¼

f f ¼ , hþf n

where n ¼ h þ f. It follows that p þ q ¼ 1, p ¼ 1  q, and q ¼ 1  p. The odds in favor of the occurrence of the event are h: f or h/f; the odds against its happening are f :h or f/h. If p is the probability that an event will occur, the odds in favor of its happening are p:q ¼ p:(1  p) or p=(1  p); the odds against its happening are q:p ¼ (1  p):p or (1  p)=p.

27.2

COMPOUND PROBABILITY

Two or more events are said to be independent if the occurrence or non-occurrence of any one of them does not affect the probabilities of occurrence of any of the others. Thus if a coin is tossed four times and it turns up a head each time, the fifth toss may be head or tail and is not influenced by the previous tosses. The probability that two or more independent events will happen is equal to the product of their separate probabilities. Thus the probability of getting a head on both the fifth and sixth tosses is 12 ( 12 ) ¼ 14. Two or more events are said to be dependent if the occurrence or non-occurrence of one of the events affects the probabilities of occurrence of any of the others. Consider that two or more events are dependent. If p1 is the probability of a first event, p2 the probability that after the first has happened the second will occur, p3 the probability that after the first and second have happened the third will occur, etc., then the probability that all events will happen in the given order is the product p1  p2  p3   . 310

CHAP. 27]

PROBABILITY

311

For example, a box contains 3 white balls and 2 black balls. If a ball is drawn at random, the probability 2 2 ¼ . If this ball is not replaced and a second ball is drawn, the probability that it also that it is black is 3þ2 5   1 1 2 1 1 ¼ . Thus the probability that both will be black is is black is ¼ . 3þ1 4 5 4 10 Two or more events are said to be mutually exclusive if the occurrence of any one of them excludes the occurrence of the others. The probability of occurrence of some one of two or more mutually exclusive events is the sum of the probabilities of the individual events. EXAMPLE 27.1. exclusive so

If a die is thrown, what is the probability of getting a 5 or a 6? Getting a 5 and getting a 6 are mutually P(5 or 6) ¼ P(5) þ P(6) ¼

1 1 2 1 þ ¼ ¼ 6 6 6 3

Two events are said to be overlapping if the events have at least one outcome in common, hence they can happen at the same time. The probability of occurrence of some one of two overlapping events is the sum of the probabilities of the two individual events minus the probability of their common outcomes.

EXAMPLE 27.2. If a die is thrown, what is the probability of getting a number less than 4 or an even number? The numbers less than 4 on a die are 1, 2, and 3. The even numbers on a die are 2, 4, and 6. Since these two events have a common outcome, 2, they are overlapping events. P(less than 4 or even) ¼ P(less than 4) þ P(even)  P(less than 4 and even) 3 3 1 þ  6 6 6 5 ¼ 6 ¼

27.3

MATHEMATICAL EXPECTATION

If p is the probability that a person will receive a sum of money m, the value of his expectation is p  m. Thus if the probability of your winning a $10 prize is 1/5, your expectation is 15 ($10) ¼ $2. 27.4

BINOMIAL PROBABILITY

If p is the probability that an event will happen in any single trial and q ¼ 1  p is the probability that it will fail to happen in any single trial, then the probability of its happening exactly r times in n trials is n Cr pr qnr . (See Problems 27.22 and 27.23.) The probability that an event will happen at least r times in n trials is pn þ n C1 pn1 q þ n C2 pn2 q2 þ    þ n Cr pr qnr : This expression is the sum of the first n  r þ 1 terms of the binomial expansion of ( p þ q)n . (See Problems 27.24 –27.26.)

27.5

CONDITIONAL PROBABILITY

The probability that a second event will occur given that the first event has occurred is called conditional probability. To find the probability that the second event will occur given that the first event occurred, divide the

312

PROBABILITY

[CHAP. 27

probability that both events occurred by the probability of the first event. The probability of event B given that event A has occurred is denoted by P(BjA). EXAMPLE 27.3. A box contains black chips and red chips. A person draws two chips without replacement. If the probability of selecting a black chip and a red chip is 15/56 and the probability of drawing a black chip on the first draw is 3/4, what is the probability of drawing a red chip on the second draw, if you know the first chip drawn was black? If B is the event drawing a black chip and R is the event drawing a red chip, then P(RjB) is the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw. P(RjB) ¼ ¼

P(R and B) P(B) 15=56 3=4

15 4  56 3 5 ¼ 14 ¼

Thus, the probability of drawing a red chip on the second draw given that a black chip was drawn on the first draw is 5/14.

Solved Problems 27.1

One ball is drawn at random from a box containing 3 red balls, 2 white balls, and 4 blue balls. Determine the probability p that it is (a) red, (b) not red, (c) white, (d ) red or blue. SOLUTION (a) p ¼

ways of drawing 1 out of 3 red balls 3 3 1 ¼ ¼ ¼ ways of drawing 1 out of (3 þ 2 þ 4) balls 3 þ 2 þ 4 9 3

(b) p ¼ 1 

27.2

1 2 ¼ 3 3

(c) p ¼

2 9

(d ) p ¼

3þ4 7 ¼ 9 9

One bag contains 4 white balls and 2 black balls; another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, determine the probability p that (a) both are white, (b) both are black, (c) 1 is white and 1 is black. SOLUTION    4 3 1 ¼ (a) p ¼ 4þ2 3þ5 4



  2 5 5 ¼ 4þ2 3þ5 24   4 5 5 ¼ . (c) Probability that first ball is white and second black ¼ 6 8 12   2 3 1 Probability that first ball is black and second white ¼ ¼ . 6 8 8 (b) p ¼

These are mutually exclusive; hence the required probability p ¼ Another method.

27.3

5 1 13 þ ¼ . 12 8 24

1 5 13 p¼1  ¼ . 4 24 24

Determine the probability of throwing a total of 8 in a single throw with two dice, each of whose faces is numbered from 1 to 6.

CHAP. 27]

313

PROBABILITY

SOLUTION Each of the faces of one die can be associated with any of the 6 faces of the other die; thus the total number of possible cases ¼ 6 . 6 ¼ 36 cases. There are 5 ways of throwing an 8: 2, 6; 3, 5; 4, 4; 5, 3; 6, 2. Required probability ¼

27.4

number of favorable cases 5 ¼  possible number of cases 36

What is the probability of getting at least 1 one in 2 throws of a die? SOLUTION The probability of not getting a one in any single throw ¼ 1 2 1/6 ¼ 5/6. The probability of not getting a one in 2 throws ¼ (5/6)(5/6) ¼ 25/36. Hence the probability of getting at least 1 one in 2 throws ¼ 1 2 25/36 ¼ 11/36.

27.5

The probability of A’s winning a game of chess against B is 1/3. What is the probability that A will win at least 1 of a total of 3 games? SOLUTION The probability of A’s losing any single game ¼ 1 2 1/3 ¼ 2/3, and the probability of A losing all 3 games ¼ (2/3)3 ¼ 8/27. Hence the probability of A winning at least 1 game ¼ 1 2 8/27 ¼ 19/27.

27.6

Three cards are drawn from a pack of 52, each card being replaced before the next one is drawn. Compute the probability p that all are (a) spades, (b) aces, (c) red cards. SOLUTION A pack of 52 cards includes 13 spades, 4 aces, and 26 red cards.  3  3  3 13 1 4 1 26 1 (b) p ¼ (c) p ¼ (a) p ¼ ¼ ¼ ¼ 52 64 52 2197 52 8

27.7

The odds are 23 to 2 against a person winning a $500 prize. What is her mathematical expectation? SOLUTION

 2 Expectation ¼ probability of winning  sum of money ¼ ($500) ¼ $40: 23 þ 2 27.8



Nine tickets, numbered from 1 to 9, are in a box. If 2 tickets are drawn at random, determine the probability p that (a) both are odd, (b) both are even, (c) one is odd and one is even, (d ) they are numbered 2, 5. SOLUTION There are 5 odd and 4 even number tickets. (a) p ¼ (b) p ¼

number of selections of 2 out of 5 odd tickets 5 C2 5 ¼ ¼ number of selections of 2 out of 9 tickets 18 9 C2 4 C2 9 C2

¼

1 6

(c) p ¼

5 C1  4 C1 9 C2

¼

54 5 ¼ 36 9

(d ) p ¼

2 C2 9 C2

¼

1 36

314

27.9

PROBABILITY

[CHAP. 27

A bag contains 6 red, 4 white, and 8 blue balls. If 3 balls are drawn at random, determine the probability p that (a) all 3 are red, (b) all 3 are blue, (c) 2 are white and 1 is red, (d ) at least 1 is red, (e) 1 of each color is drawn, ( f ) the balls are drawn in the order red, white, blue. SOLUTION (a) p ¼ (b) p ¼ (c) p ¼

number of selections of 3 out of 6 red balls 5 6 C3 ¼ ¼ C number of selections of 3 out of 18 balls 204 18 3 8 C3 18 C3

¼

7 102

 6 C1 3 ¼ 68 C 18 3

4 C2

(4 þ 8)C3 12 C3 55 : ¼ ¼ C C 204 18 3 18 3 55 149 ¼ : Hence the probability that at least 1 is red ¼ 1  204 204 648 648 4 ¼ ¼ (e) p ¼ 18  17  16=6 17 18 C3 (d ) Probability that none is red ¼

(f) p ¼

4 1 4 1 2  ¼ : ¼ 17 3! 17 6 51

or

p¼

648 648 2 ¼ ¼ 18  17  16 51 18 P3

27.10 Three cards are drawn from a pack of 52 cards. Determine the probability p that (a) all are aces, (b) all are aces and drawn in the order spade, club, diamond, (c) all are spades, (d ) all are of the same suit, (e) no two are of the same suit. SOLUTION (a) There are

52 C3

p¼

Hence (b) There are

52 P3

Hence

13 C3

4 C3 52 C3

¼

4 C1 52 C3

¼

1 : 5525

orders of drawing 3 out of 52 cards, one of which is the given order. p¼

Hence (c) There are

selections of 3 out of 52 cards, and 4 C3 selections of 3 out of 4 aces.

1 52 P3

¼

1 1 ¼ : 52  51  50 132 600

selections of 3 out of 13 spades. p¼

13 C3 52 C3

¼

11 : 850

(d ) There are 4 suits, each consisting of 13 cards. Hence there are 4 ways of selecting a suit, and selecting 3 cards from a given suit. Hence

p¼

13 C3

ways of

4  13 C3 22 : ¼ C 425 52 3

(e) There are 4 C3 ¼ 4 C1 ¼ 4 ways of selecting 3 out of 4 suits, and 13  13  13 ways of selecting 1 card from each of 3 given suits. Hence

p¼

4  13  13  13 169 ¼ : 425 52 C3

CHAP. 27]

315

PROBABILITY

27.11 What is the probability that any two different cards of a well-shuffled deck of 52 cards will be together in the deck, if their suit is not considered? SOLUTION Consider the probability that, for example, an ace and a king are together. There are 4 aces and 4 kings in a deck. Hence an ace can be chosen in 4 ways, and when that is done a king can be chosen in 4 ways. Thus an ace and then a king can be selected in 4  4 ¼ 16 ways. Similarly, a king and then an ace can be selected in 16 ways. Then an ace and a king can be together in 2  16 ¼ 32 ways. For every one way the combination (ace, king) occurs, the remaining 50 cards and the (ace, king) combination can be permuted in 51! ways. The number of favorable arrangements is thus 32(51!). Since the total number of arrangements of all the cards in the deck is 52!, the required probability is 32(51!) 32 8 ¼ ¼ : 52! 52 13

27.12 A man holds 2 of a total of 20 tickets in a lottery. If there are 2 winning tickets, determine the probability that he has (a) both, (b) neither, (c) exactly one. SOLUTION (a) There are

20 C2

ways of selecting 2 out of 20 tickets.

Hence the probability of his winning both prizes ¼

1 1 ¼ : 190 20 C2

Another method. The probability of winning the first prize ¼ 2/20 ¼ 1/10. After winning the first prize (he has 1 ticket left and there remain 19 tickets from which to choose the second prize) the probability of winning the second prize is 1/19.   1 1 1 ¼ : Hence the probability of winning both prizes ¼ 10 19 190 (b) There are 20 tickets, 18 of which are losers. 153 18 C2 Hence the probability of winning neither prize ¼ ¼ : 190 C 20 2 Another method. The probability of not winning the first prize ¼ 1  2=20 ¼ 9=10. If he does not win the first prize (he still has 2 tickets), the probability of not winning the second prize ¼ 1  2=19 ¼ 17=19:   9 17 153 ¼ : Hence the probability of winning neither prize ¼ 10 19 190 (c) Probability of winning exactly one prize ¼ 1  probability of winning neither  probability of winning both ¼1

153 1 36 18  ¼ ¼ : 190 190 190 95

Another method.

  2 18 9 ¼ : 20 19 95   18 2 9 ¼ : Probability of not winning first but winning second ¼ 20 19 95 Probability of winning first but not second prize ¼

Hence the probability of winning exactly 1 prize ¼

9 9 18 þ ¼ : 95 95 95

27.13 A box contains 7 tickets, numbered from 1 to 7 inclusive. If 3 tickets are drawn from the box, one at a time, determine the probability that they are alternately either odd, even, odd or even, odd, even.

316

PROBABILITY

[CHAP. 27

SOLUTION The probability that the first drawn is odd (4/7), then the second even (3/6) and then the third odd (3/5) is    4 3 3 6 ¼ : 7 6 5 35 The probability that the first drawn is even (3/7), then the second odd (4/6), and then the third even (2/5) is    3 4 2 4 ¼ : 7 6 5 35 6 4 2 ¼ : Hence the required probability ¼ þ 35 35 7 Another method. Possible orders of 7 numbers taken 3 at a time ¼ 7 P3 ¼ 7  6  5 ¼ 210: Orders where numbers are alternately odd, even, odd ¼ 4  3  3 ¼ 36. Orders where numbers are alternately even, odd, even ¼ 3  4  2 ¼ 24: Hence the required probability ¼

36 þ 24 60 2 ¼ ¼ : 210 210 7

27.14 The probability that A can solve a given problem is 4/5, that B can solve it is 2/3, and that C can solve it is 3/7. If all three try, compute the probability that the problem will be solved. SOLUTION The probability that A will fail to solve it ¼ 1 2 4/5 ¼ 1/5, that B will fail ¼ 1 2 2/3 ¼ 1/3, and that C will fail ¼ 1 2 3/7 ¼ 4/7.    1 1 4 The probability that all three fail ¼ : 5 3 7 Hence the probability that all three will not fail, i.e., that at least one will solve it, is 1

   1 1 4 4 101 ¼1 ¼ : 5 3 7 105 105

27.15 The probability that a certain man will be alive 25 years hence is 3/7, and the probability that his wife will be alive 25 years hence is 4/5. Determine the probability that, 25 years hence, (a) both will be alive, (b) at least one of them will be alive, (c) only the man will be alive. SOLUTION

  3 4 12 ¼ : 7 5 35      3 4 4 1 4 1 ¼ ¼ . (b) The probability that both will die within 25 years ¼ 1  7 5 7 5 35 (a) The probability that both will be alive ¼

Hence the probability that at least one will be alive ¼ 1 

4 31 ¼ . 35 35

(c) The probability that the man will be alive ¼ 3/7, and the probability that his wife will not be alive ¼   1  4=5 ¼ 1=5. 3 1 3 Hence the probability that only the man will be alive ¼ ¼ . 7 5 35

27.16 There are three candidates, A, B, and C, for an office. The odds that A will win are 7 to 5, and the odds that B will win are 1 to 3. (a) What is the probability that either A or B will win? (b) What are the odds in favor of C? SOLUTION (a) Probability that A will win:

7 7 ¼ , 7 þ 5 12

CHAP. 27]

PROBABILITY

that B will win:

317

1 1 ¼ . 1þ3 4

Hence the probability that either A or B will win ¼ (b) Probability that C will win: 1 

7 1 5 þ ¼ . 12 4 6

5 1 ¼ . 6 6

Hence the odds in favor of C are 1 to 5.

27.17 One purse contains 5 dimes and 2 quarters, and a second purse contains 1 dime and 3 quarters. If a coin is taken from one of the two purses at random, what is the probability that it is a quarter? SOLUTION The probability of selecting the first purse (1/2) and of then drawing a quarter from it (2/7) is (1/2) (2/7) ¼ 1/7. The probability of selecting the second purse (1/2) and of then drawing a quarter from it (3/4) is (l/2) (3/4) ¼ 3/8. Hence the required probability ¼

1 3 29 þ ¼ . 7 8 56

27.18 A bag contains 2 white balls and 3 black balls. Four persons, A, B, C, D, in the order named, each draw one ball and do not replace it. The first to draw a white ball receives $10. Determine their expectations. SOLUTION 2 2 A’s probability of winning ¼ , and the expectation ¼ ($10) ¼ $4. 5 5 To find B’s expectation: The probability that A fails ¼ 1 2 2/5 ¼ 3/5. If A fails, the bag contains 2 white and 2 black balls. Thus the probability that if A fails B will win ¼ 2/4 ¼ 1/2. Hence B’s probability of winning ¼ (3/5)(1/2) ¼ 3/10, and the expectation is $3. To find C ’s expectation: The probability that A fails ¼ 3/5, and the probability that B fails ¼ 1 2 1/2 ¼ 1/2. If A and B both fail, the bag contains 2 white balls and 1 black ball. Thus the probability that, if A and B both fail, C    3 1 2 1 1 will win ¼ 2/3. Hence C ’s probability of winning ¼ ¼ , and the expectation ¼ ($10) ¼ $2. 5 2 3 5 5 If A, B, and C fail, only white balls remain and D must win. Hence D’s probability of     3 1 1 1 1 1 winning ¼ ¼ , and the expectation ¼ ($10) ¼ $1. 5 2 3 1 10 10 2 3 1 1 Check. $4 þ $3 þ $2 þ $1 ¼ $10, and þ þ þ ¼ 1. 5 10 5 10

27.19 Eleven books, consisting of 5 engineering books, 4 mathematics books, and 2 chemistry books, are placed on a shelf at random. What is the probability p that the books of each kind are all together? SOLUTION When the books of each kind are all together, the engineering books could be arranged in 5! ways, the mathematics books in 4! ways, the chemistry books in 2! ways, and the 3 groups in 3! ways. p¼

ways in which books of each kind are together 5!4!2!3! 1 ¼ ¼ : total number of ways of arranging 11 books 11! 1155

27.20 Five red blocks and 4 white blocks are placed at random in a row. What is the probability p that the extreme blocks are both red? SOLUTION Total possible arrangements of 5 red and 4 white blocks ¼

(5 þ 4)! 9! ¼ ¼ 126. 5!4! 5!4!

318

PROBABILITY

[CHAP. 27

(9  2)! 7! ¼ ¼ 35. Arrangements where extreme blocks are both red ¼ (5  2)!4! 3!4! 35 5 Hence the required probability p ¼ ¼ . 126 18

27.21 One purse contains 6 copper coins and 1 silver coin; a second purse contains 4 copper coins. Five coins are drawn from the first purse and put into the second, and then 2 coins are drawn from the second and put into the first. Determine the probability that the silver coin is in (a) the second purse, (b) the first purse. SOLUTION Initially, the first purse contains 7 coins. When 5 coins are drawn from the first purse and put into the second, the probability that the silver coin is put into the second purse is 5/7, and the probability that it remains in the first purse is 2/7. The second purse now contains 5 þ 4 ¼ 9 coins. Finally, after 2 of these 9 coins are put into the first purse,   5 7 5 the probability that the silver coin is in the second purse ¼ ¼ , and the probability that it is in the first 7 9 9     2 5 2 4 5 4 ¼ or 1  ¼ . purse ¼ þ 7 7 9 9 9 9

27.22 Compute the probability that a single throw with 9 dice will result in exactly 2 ones. SOLUTION

   2 1 1 1 . The probability that the The probability that a certain pair of the 9 dice thrown will yield ones ¼ ¼ 6 6 6  7  7 1 5 other 7 dice will not yield ones ¼ 1  ¼ . Since 9C2 different pairs may be selected from the 9 dice, the 6 6  2  7 1 5 78 125 probability that exactly 1 pair will be aces is 9 C2 . ¼ 6 6 279 936  2  7 1 5 78 125 : Or, by formula: Probability ¼ n Cr pr qnr ¼ 9 C2 ¼ 6 6 279 936

27.23 What is the probability of getting a 9 exactly once in 3 throws with a pair of dice? SOLUTION A 9 can occur in 4 ways: 3, 6; 4, 5; 5, 4; 6, 3. 1 In any throw with a pair of dice, the probability of getting a 9 ¼ 4=(6  6) ¼ , and the probability of not getting 9 1 8 a 9 ¼ 1  ¼ . The probability that any given throw with a pair of dice is a 9 and that the other two throws are  9 92 1 8 . Since there are 3C1 ¼ 3 different ways in which one throw is a 9 and the other two throws are not, not ¼ 9 9   2 1 8 64 ¼ . the probability of throwing a 9 exactly once in 3 throws ¼ 3 C1 9 9 243   2 1 8 64 r nr ¼ 3 C1 ¼ . Or, by formula: Probability ¼ n Cr p q 9 9 243

27.24 If the probability that the average freshman will not complete four years of college is 1/3, what is the probability p that of 4 freshmen at least 3 will complete four years of college? SOLUTION Probability that 3 will complete and 1 will not ¼ 4 C3

 3    3   2 1 2 1 ¼ 4 C1 . 3 3 3 3

CHAP. 27]

319

PROBABILITY

 4  3   2 2 1 16 þ 4 C1 ¼ . 3 3 3 27   2 1 4 þ Or, by formula: p ¼ first 2 (n  r þ 1 ¼ 4  3 þ 1) terms of the expansion of 3 3  4  3   2 2 1 16 32 16 ¼ þ 4 C1 þ ¼ : ¼ 3 3 3 81 81 27 Probability that 4 will complete ¼

 4 2 . 3

Hence p ¼

27.25 A coin is tossed 6 times. What is the probability p of getting at least 3 heads? What are the odds in favor of getting at least 3 heads? SOLUTION On each toss, probability of a head ¼ probability of a tail ¼ 1/2. The probability that certain 3 of the 6 tosses will give heads ¼ (1/2)3. The probability that none of the other 3 tosses will be heads ¼ (1/2)3. Since 6C3, different selections of 3 can be made from the 6 tosses, the probability that exactly 3 will be heads is

6 C3

 3  3  6 1 1 1 ¼ 6 C3 : 2 2 2

Similarly, the probability of exactly 4 heads ¼ 6C4(1/2)6 ¼ 6C2(1/2)6, the probability of exactly 5 heads ¼ 6C5(1/2)6 ¼ 6C1(1/2)6, the probability of exactly 6 heads ¼ (1/2)6.  6  6  6  6 1 1 1 1 Hence p¼ þ 6 C1 þ 6 C2 þ 6 C3 2 2 2 2  6  6 1 1 21 ¼ (1 þ 6 C1 þ 6 C2 þ 6 C3 ) ¼ (1 þ 6 þ 15 þ 20) ¼ : 2 2 32 The odds in favor of getting at least 3 heads is 21:11 or 21/11.  Or, by formula: p ¼ first 4 (n  r þ 1 ¼ 6  3 þ 1) terms of the expansion of ¼

 1 1 6 þ 2 2

 6  6  6  6 1 1 1 1 21 þ 6 C1 þ 6 C2 þ 6 C3 ¼ : 2 2 2 2 32

27.26 Determine the probability p that in a family of 5 children there will be at least 2 boys and 1 girl. Assume that the probability of a male birth is 1/2. SOLUTION The three favorable cases are: 2 boys, 3 girls; 3 boys, 2 girls; 4 boys, 1 girl. p¼

 5 1 1 25 (5 C2 þ 5 C3 þ 5 C4 ) ¼ (10 þ 10 þ 5) ¼ : 2 32 32

27.27 The probability that a student takes chemistry and is on the honor rolls is 0.042. The probability that a student is on the honor roll is 0.21. What is the probability that the student is taking chemistry, given that the student is on the honor roll?

320

PROBABILITY

[CHAP. 27

SOLUTION P(taking chemistryjon honor roll) ¼ ¼

P(taking chemistry and on honor roll) P(on honor roll) 0:042 0:21

¼ 0:2

27.28 At the Pine Valley Country Club, 32% of the members play golf and are female. Also, 80% of the members play golf. If a member of the club is selected at random, find the probability that the member is female given that the member plays golf. SOLUTION P(femalejgolfer) ¼ ¼

P(female and golfer) P(golfer) 0:32 0:8

¼ 0:4

Supplementary Problems 27.29

Determine the probability that a digit chosen at random from the digits 1, 2, 3, . . . , 9 will be (a) odd, (b) even, (c) a multiple of 3.

27.30

A coin is tossed three times. If H ¼ head and T ¼ tail, what is the probability of the tosses coming up in the order (a) HTH, (b) THH, (c) HHH?

27.31

If three coins are tossed, what is the probability of obtaining (a) three heads, (b) two heads and a tail?

27.32

Find the probability of throwing a total of 7 in a single throw with two dice.

27.33

What is the probability of throwing a total of 8 or 11 in a single throw with two dice?

27.34

A die is thrown twice. What is the probability of getting a 4 or 5 on the first throw and a 2 or 3 on the second throw? What is the probability of not getting a one on either throw?

27.35

What is the probability that a coin will turn up heads at least once in six tosses of a coin?

27.36

Five discs in a bag are numbered 1, 2, 3, 4, 5. What is the probability that the sum of the numbers on three discs chosen at random is greater than 10?

27.37

Three balls are drawn at random from a box containing 5 red, 8 black, and 4 white balls. Determine the probability that (a) all three are white, (b) two are black and one red, (c) one of each color is selected.

27.38

Four cards are drawn from a pack of 52 cards. Find the probability that (a) all are kings, (b) two are kings and two are aces, (c) all are of the same suit, (d ) all are clubs.

27.39

A woman will win $3.20 if in 5 tosses of a coin she gets either of the sequences HTHTH or THTHT where H ¼ head and T ¼ tail. Determine her expectation.

27.40

In a plane crash it was reported that three persons out of the total of twenty passengers were injured. Three newspapermen were in this plane. What is the probability that the three reported injured were the newspapermen?

CHAP. 27]

PROBABILITY

321

27.41

A committee of three is to be chosen from a group consisting of 5 men and 4 women. If the selection is made at random, find the probability that (a) all three are women, (b) two are men.

27.42

Six persons seat themselves at a round table. What is the probability that two given persons are adjacent?

27.43

A and B alternately toss a coin. The first one to turn up a head wins. If no more than five tosses each are allowed for a single game, find the probability that the person who tosses first will win the game. What are the odds against A’s losing if she goes first?

27.44

Six red blocks and 4 white blocks are placed at random in a row. Find the probability that the two blocks in the middle are of the same color.

27.45

In 8 tosses of a coin determine the probability of (a) exactly 4 heads, (b) at least 2 tails, (c) at most 5 heads, (d ) exactly 3 tails.

27.46

In 2 throws with a pair of dice determine the probability of getting (a) an 11 exactly once, (b) a 10 twice.

27.47

What is the probability of getting at least one 11 in 3 throws with a pair of dice?

27.48

In ten tosses of a coin, what is the probability of getting not less than 3 heads and not more than 6 heads?

27.49

The probability that an automobile will be stolen and found within one week is 0.0006. The probability that an automobile will be stolen is 0.0015. What is the probability that a stolen automobile will be found in one week?

27.50

In the Pizza Palace, 95% of the customers order pizza. If 65% of the customers who order pizza also order breadsticks, find the probability that a customer who orders a pizza will also order breadsticks.

27.51

In a large shopping mall, a marketing agency conducted a survey of 100 people about a ban on smoking in the mall. Of the 60 non-smokers surveyed, 48 preferred a smoking ban. Of the 40 smokers surveyed, 32 preferred a smoking ban. What is the probability that a person selected at random from the group surveyed prefers a smoking ban given that the person is a non-smoker?

27.52

In a new subdivision, 35% of the houses have a family room and a fireplace, while 70% have family rooms. What is the probability that a house selected at random in this subdivision has a fireplace given that it has a family room?

ANSWERS TO SUPPLEMENTARY PROBLEMS 27.29

(a) 5/9

(b) 4/9

(c) 1/3

27.30

(a) 1/8

(b) 1/8

(c) 1/8

27.31

(a) 1/8

(b) 3/8

27.32

1/6

27.33

7/36

27.34

1/9, 25/36

27.35

63/64

27.36

1/5

27.37

(a)

1 170

(b)

7 34

(c)

4 17

322

PROBABILITY

1 270 725

27.38

(a)

27.39

20 cents

27.40

1/1140

27.41

(a) 1/21

27.42

2/5

27.43 27.44 27.45 27.46 27.47 27.48

21 , 32 7 15

36 270 725

(c)

44 4165

(b) 10/21

21:11

35 128 17 (a) 162 919 5832 99 128 (a)

(b)

247 256 1 (b) 144 (b)

27.49

0.4

27.50

13 (about 68%) 19

27.51

0.8

27.52

0.5

(c)

219 256

(d )

7 32

(d )

11 4165

[CHAP. 27

CHAPTER 28

Determinants 28.1

DETERMINANTS OF SECOND ORDER

The symbol

  a1   a2

 b1  b2 

consisting of the four numbers a1, b1, a2, b2 arranged in two rows and two columns, is called a determinant of second order or determinant of order two. The four numbers are called elements of the determinant. By definition,  b1  ¼ a1 b2  b1 a2 : b2 

  a1   a2

   2 3    1 2  ¼ (2)(2)  (3)(1) ¼ 4 þ 3 ¼ 1:

Thus

Here the elements 2 and 3 are in the first row, the elements 21 and 22 are in the second row. Elements 2 and 21 are in the first column, and elements 3 and 22 are in the second column. A determinant is a number. A determinant of order one is the number itself. 28.2

CRAMER’S RULE

Systems of two linear equations in two unknowns may be solved by use of second-order determinants. Given the system of equations

a1 x þ b1 y ¼ c1 a2 x þ b2 y ¼ c2

(1)

it is possible by any of the methods of Chapter 15 to obtain x¼

c 1 b2  b1 c 2 , a1 b2  b1 a2

y¼

a1 c 2  c 1 a2 a1 b2  b1 a2 323

(a1 b2  b1 a2 = 0):

324

DETERMINANTS

[CHAP. 28

These values for x and y may be written in terms of second-order determinants as follows:   c1   c2  x¼  a1  a2

 b1  b2  , b1  b2 

  a1   a2  y¼  a1  a2

 c1  c2   b1  b2 

(2)

The form involving determinants is easy to remember if one keeps in mind the following: (a) The denominators in (2) are given by the determinant    a1 b1     a2 b2  in which the elements are the coefficients of x and y arranged as in the given equations (1). This determinant, usually denoted by D, is called the determinant of the coefficients. (b) The numerator in the solution for either unknown is the same as the determinant of the coefficients D with the exception that the column of coefficients of the unknown to be determined is replaced by the column of constants on the right side of equations (1). When the column of coefficients for the variable x in determinant D is replaced with the column of constants, we call the new determinant Dx. When the column of y coefficients in determinant D is replaced with the column of constants, we call the new determinant Dy.

2x þ 3y ¼ 8 x  2y ¼ 3:   2 3  The denominator for both x and y is D ¼  ¼ 2(2)  3(1) ¼ 7. 1 2       8 2 3  8    ¼ 8(2)  3(3) ¼ 7, Dy ¼  ¼ 2(3)  8(1) ¼ 14 Dx ¼  3 2  1 3  EXAMPLE 28.1.

Solve the system

x¼

Dx 7 ¼ ¼ 1, D 7

y¼

Dy 14 ¼ ¼2 D 7

Thus, the solution of the system is (1, 2).

The method of solution of linear equations by determinants is called Cramer’s Rule. If the determinant D ¼ 0, then Cramer’s Rule can not be used to solve the system.

28.3

DETERMINANTS OF THIRD ORDER

The symbol   a1   a2   a3

b1 b2 b3

 c1  c2  c3 

consisting of nine numbers arranged in three rows and three columns is called a determinant of third order. By definition, the value of this determinant is given by a 1 b2 c 3 þ b1 c 2 a3 þ c 1 a2 b3  c 1 b2 a3  a1 c 2 b3  b1 a2 c 3 and is called the expansion of the determinant.

CHAP. 28]

325

DETERMINANTS

In order to remember this definition, the following scheme is given. Rewrite the first two columns on the right of the determinant as follows: a1 b1 c1 a1 b1 a2 b2 c2 a2 b2 a3 b3 c3 a3 b3 – – – + + + (a) Form the products of the elements in each of the 3 diagonals shown which run down from left to right, and precede each of these 3 terms by a positive sign. (b) Form the products of the elements in each of the 3 diagonals shown which run down from right to left, and precede each of these 3 terms by a negative sign. (c) The algebraic sum of the six products of (1) and (2) is the required expansion of the determinant. EXAMPLE 28.2.    3 2 2    6 1 1    2 3 2

Evaluate Rewriting,

2 3 –2 1 –1 6 1 –2 –3 2 –2 –3 – – – + + + 3 6

–2

The value of the determinant is (3)(1)(2) þ (2)(1)(2) þ (2)(6)(3)  (2)(1)(2)  (3)(1)(3)  (2)(6)(2) ¼ 15:

Cramer’s Rule for linear equations in 3 unknowns is a method of solving the following equations for x, y, z 8 < a1 x þ b1 y þ c1 z ¼ d1 a x þ b2 y þ c2 z ¼ d2 ð3Þ : 2 a3 x þ b3 y þ c3 z ¼ d3 by determinants. It is an extension of Cramer’s Rule for linear equations in two unknowns. If we solve equations (3) by the methods of Chapter 12, we obtain x¼

d1 b2 c3 þ c1 d2 b3 þ b1 c2 d3  c1 b2 d3  b1 d2 c3  d1 c2 b3 a1 b2 c3 þ b1 c2 a3 þ c1 a2 b3  c1 b2 a3  b1 d2 c3  a1 c2 b3

y¼

a1 d2 c3 þ c1 a2 d3 þ d1 c2 a3  c1 d2 a3  d1 a2 c3  a1 c2 d3 a1 b2 c3 þ b1 c2 a3 þ c1 a2 b3  c1 b2 a3  b1 d2 c3  a1 c2 b3

z¼

a1 b2 d3 þ d1 a2 b3 þ b1 d2 a3  d1 b2 a3  b1 a2 d3  a 1 d2 b3 a1 b2 c3 þ b1 c2 a3 þ c1 a2 b3  c1 b2 a3  b1 a2 c3  a1 c2 b3

These may be written in terms of determinants as follows   a1   D ¼  a2   a3

b1 b2 b3

 c1   c2   c3 

  d1   D x ¼  d2   d3

b1 b2

x¼

Dx , D

b3

 c1   c2   c3  y¼

  a1   D y ¼  a2   a3 Dy , D

z¼

d1 d2 d3 Dz D

 c1   c2   c3 

  a1   D z ¼  a2   a3

b1 b2 b3

 d1   d2   d3  (4)

326

DETERMINANTS

[CHAP. 28

D is the determinant of the coefficients of x, y, z in equations (3) and is assumed not equal to zero. If D is zero, Cramer’s Rule cannot be used to solve the system of equations. The form involving determinants is easy to remember if one keeps in mind the following: (a) The denominators in (4) are given by the determinant D in which the elements are the coefficients of x, y, and z arranged as in the given equations (3). (b) The numerator in the solution for any unknown is the same as the determinant of the coefficients D with the exception that the column of coefficients of the unknown to be determined is replaced by the column of constants on the right side of equations (3). Dx Dy Dz (c) The solution of the system is (x, y, z) where x ¼ , y¼ , and z ¼ : D D D EXAMPLE 28.3.

Solve the system 8 < x þ 2y  z ¼ 3 3x þ y þ z ¼ 4 : x  y þ 2z ¼ 6:   1 2 1     D ¼ 3 1 1  ¼ 2 þ 2 þ 3 þ 1 þ 1  12 ¼ 3    1 1 2    3 2 1     Dx ¼  4 1 1  ¼ 6 þ 12 þ 4 þ 6  3  16 ¼ 3    6 1 2    1 3 1      Dy ¼  3 4 1  ¼ 8  3  18 þ 4  6 þ 18 ¼ 3   1 6 2   1 2 3     Dz ¼  3 1 4  ¼ 6 þ 8 þ 9 þ 3 þ 4  36 ¼ 6    1 1 6 x¼

Dx 3 ¼ ¼ 1, D 3

y¼

Dy 3 ¼ ¼ 1, D 3

z¼

Dz 6 ¼ ¼2 D 3

The solution of the system is (1, 21, 2).

28.4

DETERMINANTS OF ORDER n

An nth order determinant is written   a11   a21   a31   .  ..  a

n1

a12 a22 a32 .. . an2

a13 a23 a33 .. . an3

... ... ... ...

 a1n  a2n  a3n  ..  .  ann 

In this notation each element is characterized by two subscripts, the first indicating the row in which the element appears, the second indicating the column in which the element appears. Thus a23 is the element in the 2nd row and 3rd column whereas a32 is the element in the 3rd row and 2nd column. The principal diagonal of a determinant consists of the elements in the determinant which lie in a straight line from the upper left-hand corner to the lower right-hand corner.

CHAP. 28]

28.5 I.

PROPERTIES OF DETERMINANTS Interchanging corresponding rows and columns of a determinant does not change the value of the determinant. Thus any theorem proved true for rows holds for columns, and conversely. EXAMPLE 28.4.

II.

  a11   a21   a31

  a11   a21   a31

 a13  a23  ¼ 0 a33 

  a11   a21   a31

   a31 a13   a23  ¼  a21  a11 a33 

a12 a22 a32

a32 a22 a12

 a33  a23  a13 

  a11   a21   a31

a12 a22 a32

 a11  a21  ¼ 0 a31 

If each of the elements in a row (or column) of a determinant is multiplied by the same number p, the value of the determinant is multiplied by p.   pa11   pa21   pa31

a12 a22 a32

   a11 a13   a23  ¼ p a21  a31 a33 

a12 a22 a32

 a13  a23  a33 

If each element of a row (or column) of a determinant is expressed as the sum of two (or more) terms, the determinant can be expressed as the sum of two (or more) determinants. EXAMPLE 28.9.

VII.

0 0 0

If two rows (or columns) of a determinant are identical, the value of the determinant is zero.

EXAMPLE 28.8.

VI.

 a31  a32  a33 

Interchanging any two rows (or columns) reverses the sign of the determinant.

EXAMPLE 28.7.

V.

a21 a22 a23

If each element in a row (or column) is zero, the value of the determinant is zero.

EXAMPLE 28.6.

IV.

  a13   a11 a23  ¼  a12 a33   a13

a12 a22 a32

EXAMPLE 28.5.

III.

327

DETERMINANTS

  a11 þ a011   a21 þ a0 21   a31 þ a0 31

a12 a22 a32

  a13   a11 a23  ¼  a21 a33   a31

a12 a22 a32

  a13   a011 a23  þ  a021 a33   a031

a12 a22 a32

 a13  a23  a33 

If to each element of a row (or column) of a determinant is added m times the corresponding element of any other row (or column), the value of the determinant is not changed. EXAMPLE 28.10.

  a11 þ ma12   a21 þ ma22   a31 þ ma32

a12 a22 a32

  a13   a11 a23  ¼  a21 a33   a31

a12 a22 a32

 a13  a23  a33 

These properties may be proved for the special cases of second and third order determinants by using the methods of expansion of Sections 28.2 and 28.3. For proofs of the general cases see the solved problems below.

328

28.6

DETERMINANTS

[CHAP. 28

MINORS

The minor of an element in a determinant of order n is the determinant of order n  1 obtained by removing the row and the column which contain the given element. For example, the minor of a32 in the 4th order determinant   a11   a21   a31   a41

a12 a22 a32 a42

a13 a23 a33 a43

 a14  a24  a34  a44 

is obtained by crossing out the row and column containing a32 as shown, and writing the resulting determinant of order 3, namely   a11   a21   a41

a13 a23 a43

 a14  a24 : a44 

The minor of an element is denoted by capital letters. Thus the minor corresponding to the element a32 is denoted by A32 .

28.7

VALUE OF A DETERMINANT OF ORDER n

The value of a determinant may be obtained in terms of minors as follows: (1)

Choose any row (or column).

(2)

Multiply each element in the row (or column) by its corresponding minor preceded by (1)i þ j where i þ j is the sum of the row number i and column number j. The minor of an element with the attached sign is called the cofactor of the element. Add algebraically the products obtained in (2).

(3)

For example, let us expand the determinant   a11   a21   a31   a41

a12 a22 a32 a42

a13 a23 a33 a43

 a14  a24  a34  a44 

by the elements in the third row. The minors of a31 , a32 , a33 , a34 are A31 , A32 , A33 , A34 , respectively. The sign corresponding to the element a31 is þ since (1)3þ1 ¼ (1)4 ¼ þ1. Similarly, the signs associated with the elements a32 , a33 , a34 are , þ ,  respectively. Thus the value of the determinant is a31 A31  a32 A32 þ a33 A33  a34 A34 : Property VII is useful in producing zeros in a given row or column. This property coupled with the expansion in terms of minors makes for easy determination of the value of a determinant.

28.8

CRAMER’S RULE FOR DETERMINANTS OF ORDER n

Cramer’s Rule for the solution of n simultaneous linear equations in n unknowns is exactly analogous to the rule given in Section 28.2 for the case n ¼ 2 and in Section 28.3 for n ¼ 3.

CHAP. 28]

329

DETERMINANTS

Given n linear equations in n unknowns x1 , x2 , x3 , . . . , xn a11 x1 þ a12 x2 þ a13 x3 þ    þ a1n xn ¼ r1 a21 x1 þ a22 x2 þ a23 x3 þ    þ a2n xn ¼ r2 .. .. .. .. .. . . . . . an1 x1 þ an2 x2 þ an3 x3 þ    þ ann xn ¼ rn : Let D be the determinant of the coefficients of x1 , x2 , x3 , . . . , xn ,   a11 a12 a13 . . .   a21 a22 a23 . . .  D¼ . .. ..  .. . .  a a a ... n1

n2

n3

(1)

i.e.,

 a1n  a2n  ..  .  ann 

Denote by Dk the determinant D with the kth column (which corresponds to the coefficients of the unknown xk ) replaced by the column of the coefficients on the right-hand side of (1). Then x1 ¼

D1 , D

x2 ¼

D2 , D

x3 ¼

D3 ,... D

provided D= 0:

If D= 0 there is one and only one solution. If D ¼ 0 the system of equations may or may not have solutions. Equations having no simultaneous solution are called inconsistent, otherwise they are consistent. If D ¼ 0 and at least one of the determinants D1 , D2 , . . . , Dn = 0, the given system is inconsistent. If D ¼ D1 ¼ D2 ¼ . . . ¼ Dn ¼ 0, the system may or may not be consistent. Equations having an infinite number of simultaneous solutions are called dependent. If a system of equations is dependent then D ¼ 0 and all of the determinants D1 , D2 , . . . , Dn ¼ 0. The converse, however, is not always true.

28.9

HOMOGENEOUS LINEAR EQUATIONS

If r1 , r2 , . . . , rn in equations (1) are all zero, the system is said to be homogeneous. In this case D1 ¼ D2 ¼ D3 ¼ . . . ¼ Dn ¼ 0 and the following theorem is true. Theorem: A necessary and sufficient condition that n homogeneous linear equations in n unknowns have solutions other than the trivial solution (where all the unknowns equal zero) is that the determinant of the coefficients, D ¼ 0. A system of m equations in n unknowns may or may not have simultaneous solutions. (1) (2)

If m . n, the unknowns in n of the given equations may be obtained. If these values satisfy the remaining m  n equations the system is consistent, otherwise it is inconsistent. If m , n, then m of the unknowns may be determined in terms of the remaining n  m unknowns.

Solved Problems 28.1

Evaluate the following determinants.   3 2   ¼ (3)(4)  (2)(1) ¼ 12  2 ¼ 10 (a)  1 4    3 1   ¼ (3)(2)  (1)(6) ¼ 6 þ 6 ¼ 0  (b)  6 2 

330

DETERMINANTS

 0 (c)  2

[CHAP. 28

 3  ¼ (0)(5)  (3)(2) ¼ 0  6 ¼ 6 5 

   x x2    ¼ xy2  x2 y (d )  y y2     x þ 2 2x þ 5   ¼ (x þ 2)(x  3)  (2x þ 5)(3x  1) ¼ 5x2  14x  1 (e)  3x  1 x  3  28.2

(a) Show that if the rows and columns of a determinant of order two are interchanged the value of the determinant is the same. (b) Show that if the elements of one row (or column) are proportional respectively to the elements of the other row (or column), the determinant is equal to zero. SOLUTION

 a (a) Let the determinant be  1 a2

 b1  ¼ a1 b2  a2 b1 : b2 

The determinant withrows and  columns interchanged so that 1st row becomes 1st column and 2nd row  a1 a2   ¼ a1 b2  a2 b1 : becomes 2nd column is  b1 b2     a1 b1   ¼ a1 kb1  b1 ka1 ¼ 0: (b) The determinant with proportional rows is  ka1 kb1 

28.3

  2x  1 Find the values of x for which  xþ1

 2x þ 1  ¼ 0: 4x þ 2 

SOLUTION    2x  1 2x þ 1  2    x þ 1 4x þ 2  ¼ ð2x  1Þð4x þ 2Þ  ð2x þ 1Þðx þ 1Þ ¼ 6x  3x  3 ¼ 0: Then 2x2  x  1 ¼ (x  1)(2x þ 1) ¼ 0 so that x ¼ 1, 1=2.

28.4

Solve for the unknowns in each of the following systems. 4x þ 2y ¼ 5 (a) 3x  4y ¼ 1 SOLUTION

 4 D ¼  3

 2  ¼ 22, 4  x¼

Dx 22 ¼ ¼ 1, D 22

The solution of the system is (1, 1/2).

(b)

3u þ 2v ¼ 18 5u  v ¼ 12

  5 2  ¼ 22, Dx ¼  1 4  y¼

  4 5  ¼ 11 Dy ¼  3 1

Dy 11 1 ¼ ¼ D 22 2

CHAP. 28]

331

DETERMINANTS

SOLUTION    3 2  ¼ þ7, D ¼  5 1  u¼

   18 2  Du ¼  ¼ 42, 12 1 

Du 42 ¼ ¼ 6, 7 D

v¼

  3 Dv ¼  5

 18  ¼ 126 12 

Dv 126 ¼ ¼ 18 7 D

The solution of the system is (6, 18).

(c)

5x  2y  14 ¼ 0 2x þ 3y þ 3 ¼ 0

SOLUTION



Rewrite as

5x  2y ¼ 14 2x þ 3y ¼ 3:

 5 D ¼  2

   14 2   ¼ 36,  Dx ¼  3 3

 2  ¼ 19, 3 x¼

Dx 36 ¼ , D 19

y¼

 5 Dy ¼  2

 14  ¼ 43 3 

Dy 43 ¼ D 19

The solution of the system is (36=19, 43=19):

28.5

Solve for x and y. 8 3x  2 7y þ 1 > > < 5 þ 10 ¼ 10 (a) > > : x þ 3  2y  5 ¼ 3 2 3   6 7   ¼ 45, D¼ 3 4  x¼

(1)

Multiply (1) by 10: 6x þ 7y ¼ 103:

(2)

Multiply (2) by 6: 3x  4y ¼ 1:

   103 7   Dx ¼  ¼ 405, 1 4  Dx 405 ¼ 9, ¼ 45 D

y¼

 6 Dy ¼  3

 103  ¼ 315 1 

Dy 315 ¼7 ¼ 45 D

The solution of the system is (9, 7). 8 2 3 > > Multiply (1) by (x þ 1)(y þ 1): 2x  3y ¼ 1: > < y þ 1  x þ 1 ¼ 0 (1) (b) > 2 3 > > : þ ¼ 0 (2) Multiply (2) by (x  7)(2y  3): 3x þ 4y ¼ 27: x  7 2y  3        2 3   1 3  2 1       D¼ ¼ 17, Dx ¼  ¼ 85, Dy ¼  ¼ 51 3 4 27 4 3 27  x¼

Dx 85 ¼ 5, ¼ 17 D

The solution of the system is (5, 3).

y¼

Dy 51 ¼3 ¼ 17 D

332

28.6

DETERMINANTS

[CHAP. 28

Solve the following systems of equations. 8 3 6 1 > >

2 3 1 > : þ ¼ x y 2    3 6    ¼ 21, D¼ 2 3

  1=6 ¼  1=2

D1=x

 6  7 ¼ , 3 2

1 D1=x 7=2 1 ¼ ¼ , ¼ x 21 6 D x¼

 1=6  7 ¼ 1=2  6

1 D1=y 7=6 1 ¼ ¼ ¼ y 21 18 D

1 1 ¼ ¼ 6, 1=x 1=6

y¼

1 1 ¼ ¼ 18 1=y 1=18

The 8 solution of the system is (6, 18). 8     3 1 8 1 > 3  8 ¼3 > > > > > < 2x 5y

> 1 4 1 4 1 > > > > : : þ ¼ 1:   ¼1 x 3 y 3y x      3 8=5  12  3=2 8=5  18   ¼ , D ¼ D ¼  1=x  1 4=3  ¼ 5 , 5 1 4=3  1 D1=x 12=5 2 ¼ ¼ , ¼ x 18=5 3 D x¼

D1=y

 3 ¼  2

   3=2 3  9 ¼ D1=y ¼  1 1  2

1 D1=y 9=2 5 ¼ ¼ ¼ y 18=5 4 D

1 1 3 ¼ ¼ , 1=x 2=3 2

y¼

1 1 4 ¼ ¼ 1=y 5=4 5

The solution of the system is (3/2, 4/5). 28.7

Evaluate each of the following determinants.    3 2 2    (a)  1 4 5   6 1 2  Repeat the first two columns:

3 –2 1 4 6 –1

2 5 2

3 –2 1 4 6 –1

(3)(4)(2) þ (2)(5)(6) þ (2)(1)(1)  (2)(4)(6)  (3)(5)(1)  (2)(1)(2) ¼ 67    1 2 3   (b)  5 3 2  ¼ 29  1 1 3     2 3 2   (c)  0 2 1  ¼ 15  1 4 0   a b c   (d )  c a b  ¼ a3 þ b3 þ c3  3abc b c a

CHAP. 28]

333

DETERMINANTS

   (x  2) (y þ 3) (z  2)    (e)  2 3 4  ¼ 11x þ 6y þ z  6  1 2 1  28.8

(a) Show that if two rows (or two columns) of a third-order determinant have their corresponding elements proportional, the value of the determinant is zero. (b) Show that if the elements of any row (or column) are multiplied by any given constant and added to the corresponding elements of any other row (or column), the value of the determinant is unchanged.

SOLUTION (a) We must show that

  a1   ka1   a3

b1 kb1 b3

 c1  kc1  ¼ 0, c3 

where the elements in the first and second rows are proportional. This is shown by expansion of the determinant. (b) Let the given determinant be   a1   a2   a3

b1 b2 b3

 c1  c2 : c3 

We must show that if k is any constant   a1   a2   a3 þ ka2

  c1   a1 c2  ¼  a2 c3 þ kc2   a3

b1 b2 b3 þ kb2

b1 b2 b3

 c1  c2  c3 

where we have multiplied each of the elements in the second row of the given determinant by k and added to the corresponding elements in the third row. The result is proved by expanding each of the determinants and showing that they are equal.

28.9

Solve the following systems of equations. 8 < 2x þ y  z ¼ 5 (a) 3x  2y þ 2z ¼ 3 : x  3y  3z ¼ 2   2 1 1     and Here D ¼  3 2 2  ¼ 42    1 3 3        5 2 2 1 1  5 1          Dx ¼  3 2 Dy ¼  3 3 Dz ¼  3 2  ¼ 42, 2  ¼ 84,       2 3 3   1 2 3  1 x¼

Dx 42 ¼ 1, ¼ 42 D

y¼

The solution of the system is (1, 2, 1).

Dy 84 ¼ 2, ¼ 42 D

z¼

1 2 3

Dz 42 ¼ 1 ¼ 42 D

 5   3  ¼ 42  2 

334

DETERMINANTS

[CHAP. 28

8 < x þ 2z ¼ 7 (b) 3x þ y ¼ 5 : 2y  3z ¼ 5

Write as

Then   7   Dx ¼  5   5

 2   1 0  ¼ 9,  2 3  0

x¼

Dx 9 ¼ ¼ 1, 9 D

8 > < x þ 0y þ 2z ¼ 7 3x þ y þ 0z ¼ 5 > : 0x þ 2y  3z ¼ 5:   1 0 2     and D ¼ 3 1 0 ¼ 9    0 2 3     1 1 7 2       Dy ¼  3 Dz ¼  3 5 0  ¼ 18,     0 5 3  0 y¼

Dy 18 ¼ 2, ¼ 9 D

z¼

0 1 2

 7   5  ¼ 27  5 

Dz 27 ¼3 ¼ 9 D

The solution of the system is (1, 2, 3).

28.10 The equations for the currents i1 , i2 , i3 in a given electrical network are 8 < 3i1  2i2 þ 4i3 ¼ 2 i1 þ 3i2  6i3 ¼ 8 : 2i1  i2  2i3 ¼ 0: Find i3 . SOLUTION

 3  1  2 i3 ¼  3 1  2

2 3 1 2 3 1

 2  8  0

22 1 ¼ ¼ 4  44 2 6  2 

28.11 Write the expansion of the determinant by using minors for row one.    a11 a12 a13     a21 a22 a23     a31 a32 a33  SOLUTION        a22 a23     þ a12 ð1Þ1þ2  a21 a23  þ a13 ð1Þ1þ3  a21 The expansion is a11 ð1Þ1þ1     a a32 a33 a31 a33 31 a11 ðþ1Þða22 a33  a23 a32 Þ þ a12 ð1Þða21 a33  a23 a31 Þ þ a13 ðþ1Þða21 a32  a22 a31 Þ ¼

 a22  ¼ a32 

a11 a22 a33  a11 a23 a32  a12 a21 a33 þ a12 a23 a31 þ a13 a21 a32  a13 a22 a31 : The required expansion is a11 a22 a33  a11 a23 a32  a12 a21 a33 þ a12 a23 a31 þ a13 a21 a32  a13 a22 a31 :

CHAP. 28]

335

DETERMINANTS

28.12 Prove Property III: If two rows (or columns) are interchanged, the sign of the determinant is changed. SOLUTION For the case of third-order determinants, we must show that

  a11    a21   a31   a31    a21   a11

  a11    a21   a31

a12 a22 a32

   a31 a13     a23  ¼  a21     a11 a33

a32 a22 a12

 a33   a23 :  a13 

a12 a22 a32

 a13    a22  a23  ¼ a11 (1)1þ1   a32 a33 

a32 a22 a12

¼ þ a11 (a22 a33  a23 a32 )  a12 (a21 a33  a23 a31 ) þ a13 (a21 a32  a22 a31 ) ¼ þ a11 a22 a33  a11 a23 a32  a12 a21 a33 þ a12 a23 a31 þ a13 a21 a32  a13 a22 a31 :  a33    !       a23  a23  a22   1þ1  a22 1þ2  a21 1þ3  a21 þ a32 (1)  þ a13 (1)  a23  ¼  a31 (1)   a11 a13  a11 a13  a11 a12  a13 

   a23  1þ2  a21 þ a (1) 12  a a33 31

   a23  1þ3  a21 þ a (1) 13  a a33 31

 a22  a32 

¼ (þa31 (a22 a13  a12 a23 )  a32 (a21 a13  a23 a11 ) þ a33 (a21 a12  a11 a22 )) ¼ a31 a22 a13 þ a31 a12 a23 þ a32 a21 a13  a32 a23 a11  a33 a21 a12 þ a33 a11 a22 ¼ þa11 a22 a33  a11 a23 a32  a12 a21 a33 þ a12 a23 a31 þ a13 a21 a32  a13 a22 a31 : The expansion of each side of the equation yields the same expression. Thus the property holds for third-order determinants. The methods of proof holds in the general case.

28.13 Prove Property IV: If two rows (or columns) are identical, the determinant has value zero. SOLUTION Let D be the value of the determinant. By Property III, interchange of the two identical rows should change the value to D. Since the determinants are the same, D ¼ D or D ¼ 0.

28.14 Prove Property V: If each of the elements of a row (or column) are multiplied by the same number p, the value of the determinant is multiplied by p. SOLUTION Each term of the determinant contains one and only one element from the row multiplied by p and thus each term has factor p. This factor is therefore common to all the terms of the expansion and so the determinant is multiplied by p.

28.15 Prove Property VI: If each element of a row (or column) of a determinant is expressed as the sum of two (or more) terms, the determinant can be expressed as the sum of two (or more) determinants. SOLUTION For the case of third-order determinants we must show that   a11   a21   a31

a011 a021 a031

a12 a22 a32

  a13   a11 a23  ¼  a21 a33   a31

a12 a22 a32

  a13   a011 a23  þ  a021 a33   a031

a12 a22 a32

 a13  a23  a33 

336

DETERMINANTS

[CHAP. 28

We will expand each determinant by minors using the first column.

  a11    a21   a31

a12 a22 a32

  a11    a21   a31  a13   a23  þ  a33 

a011

a12

a021 a031

a22 a32

 0  a11   0  a21   a0 31

a12 a22 a32

 a13   a23  ¼ (a11 þ a011 )A11  (a21 þ a021 )A21 þ (a31 þ a031 )A31 :  a33   a13   a23  ¼ a11 A11  a21 A21 þ a31 A31 þ a011 A11  a021 A21 þ a031 A31  a33  ¼ (a11 þ a011 )A11  (a21 þ a021 )A21 þ (a31 þ a031 )A31 :

The expansion for each side is identical. Thus the property holds for third-order determinants. The method of proof holds in the general case.

28.16 Prove Property VII: If to each element of a row (or column) of a determinant is added m times the corresponding element of any other row (or column), the value of the determinant is not changed. SOLUTION For the case of a third-order determinant we must show that     a11 þ ma12 a12 a13   a11     a21 þ ma22 a22 a23  ¼  a21     a31 þ ma32 a32 a33   a31 By property VI the left-hand side may be written     a11 a12 a13   ma12     a21 a22 a23  þ  ma22     a31 a32 a33   ma32 This last determinant may be written

  a12  m a22  a32

a12 a22 a32

a12 a22 a32

a12 a22 a32

 a13  a23 : a33 

 a13  a23 : a33 

 a13  a23  a33 

which is zero by Property IV.

28.17 Show that

  3 2   6 5   9 3   12 2

2 4 6 8

 1  2  ¼ 0: 5  7

SOLUTION The number 3 may be factored from each element in the first column and 2 may be factored from each element in the third column to yield   1 2 1 1   2 5 2 2  (3)(2) 3 3 3 5   4 2 4 7 which equals zero since the first and third columns are identical.

CHAP. 28]

337

DETERMINANTS

28.18 Use Property VII to transform the determinant   1   2   2

 3  4  1

2 1 3

into a determinant of equal value with zeros in the first row, second and third columns.

SOLUTION Multiply each element in the first column by 2 and add to the corresponding elements in the second column, thus obtaining   1 (2)(1)  2   2 (2)(2)  1   2 (2)(2) þ 3

   0 3  3   1 3 4 : 4  ¼  2 1   2 1 1 

Multiply each element in the first column of the new determinant by 3 and add to the corresponding elements in the third column to obtain   1   2   2

0 3 1

   0 0  (3)(1) þ 3   1 3 2 : (3)(2) þ 4  ¼  2 7 (3)(2) þ 1   2 1

The result could have been obtained in one step by writing   1 (2)(1)  2   2 (2)(2)  1   2 (2)(2) þ 3

  (3)(1) þ 3   1 0 3 (3)(2) þ 4  ¼  2 (3)(2) þ 1   2 1

 0  2 : 7

The choice of the numbers 2 and 3 was made in order to obtain zeros in the desired places.

28.19 Use Property VII to transform the determinant   3   2   4   1

6 1 5 3

2 2 1 4

 3  2  4  2 

into an equal determinant having three zeros in the 4th row.

SOLUTION Multiply each element in the 1st column (the basic column) by 3, 4, þ2 and add respectively to the corresponding elements in the 2nd, 3rd, 4th columns. The result is   3   2   4   1

(3)(3) þ 6 (3)(2) þ 1 (3)(4)  5 (3)(1) þ 3

(4)(3) þ 2 (4)(2)  2 (4)(4) þ 1 (4)(1) þ 4

  (2)(3) þ 3   3 3 7 (2)(2) þ 2   2 ¼ (2)(4) þ 4   4 17 0 (2)(1)  2   1

 10 9  6 2  15 12  0 0

Note that it is useful to choose a basic row or column containing the element 1.

338

DETERMINANTS

[CHAP. 28

28.20 Obtain 4 zeros in a row or column of the 5th order determinant   3 5   2 3   4 1   6 3   2 2

4 2 3 2 5

 6 2  3 4  2 3  4 3  3 2 

SOLUTION We shall produce zeros in the 2nd column by use of the basic row shown shaded. Multiply the elements in this basic row by 5, 3, 3, 2 and add respectively to the corresponding elements in the 1st, 2nd, 4th, 5th rows to obtain    17 0 11 16 17     14 0 7 9 13    4 1 3 2 3    18 0 11 2 6    6 0 1 7 4

28.21 Obtain 3 zeros in a row or column of the determinant   3   2   2   4

4 2 2 3 3 3 5 2

 3  2  4  2 

without changing its value. SOLUTION It is convenient to use Property VII to obtain an element 1 in a row or column. For example, by multiplying each of the elements in column 2 by 1 and adding to the corresponding elements in column 3, we obtain   3 4   2 2   2 3   4 5

 2 3  1 2  : 6 4   7 2

Using the 3rd column as the basic column, multiply its elements by 2, 2, 2 and add respectively to the 1st, 2nd, 4th columns to obtain    1 8 2 1    0 0 1 0     14 15 6 16    10 19 7 16  which equals the given determinant.

28.22 Write the minor and corresponding cofactor of the element in the second row, third column for the determinant    2 2 3 1   1 5  3 2   1 2 5 1 :   2 1 3 2 

CHAP. 28]

339

DETERMINANTS

SOLUTION Crossing out the row and column containing the element, the minor is given by    2 2 1    1 2 1 :   2 1 2  Since the element is in the 2nd row, 3rd column and 2 þ 3 ¼ 5 is an odd number, the associated sign is minus. Thus the cofactor corresponding to the given element is    2 2 1    1 2 1 : 2 1 2 

28.23 Write the minors and cofactors of the elements in the 4th row of the determinant    3 2 4 2    2 1 5 3   :  1 5 2 2    3 2 4 1 SOLUTION The elements in the 4th row are 3, 2, 4, 1.    2 4 2     Minor of element 3 ¼  1 5 3     5 2 2   3 4 2     Minor of element 2 ¼  2 5 3     1 2 2    3 2 2     Minor of element 4 ¼  2 1 3    1 5 2    3 2 4     Minor of element 1 ¼  2 1 5   1 5 2 

Cofactor ¼ Minor

Cofactor ¼ þMinor

Cofactor ¼ Minor

Cofactor ¼ þMinor

28.24 Express the value of the determinant of Problem 28.23 in terms of minors or cofactors. SOLUTION Value of determinant ¼ sum of elements each multiplied by associated cofactor 8  8  9 9   4 2 > 4 2 > > > <  2 = < 3 =     ¼ (3)  1 5 3  þ (2) þ 2 5 3  > > > > :  :  ; ; 5 2 2 1 2 2 8  8  9 9    2 > > > <  3 2 4 > <  3 2 = =     þ (4)  2 1 3  þ (1) þ 2 1 5 > > > > :  :  ; ; 1 5 2 1 5 2 Upon evaluating each of the 3rd order determinants the result 53 is obtained.

CHAP. 28]

341

DETERMINANTS

Multiply the elements in the indicated basic column by 7 and add to the corresponding elements in the 1st and 3rd columns to obtain   44    0   31   1    1  (b)  3   2   3

3 2

2

3

1 1 2 3 3 2 4

2 1 4 2

5 1 3

 37    44  0  ¼ (1) þ  31 28 

 37  ¼ 85: 28 

 1   3   4   1   1

Multiply the elements in the indicated basic column by 3, 2, 1, 4 respectively and add to the corresponding elements in the 1st, 3rd, 4th, 5th columns to obtain   8    5   0    7   3

3 2

4 5

1

0

3 2

3 8

 6 13  8   8 >  >  4 11  > <  5    0 0  ¼ 1   7 >  > > 1 11  :   3 0 9

9   8 4 6 13 > >  > =   5 5 4 11    ¼   7 3 1 11 > >  > ;   3 8 0 9

4 5 3 8

 6 13   4 11  : 1 11  0 9

In the last determinant, multiply the elements in the indicated basic row by 6, 24 and add respectively to the elements in the 1st and 2nd rows to obtain   50 22   33 17    7 3   3 8

0 0 1 0

 8  79    > <  50  55    ¼ (1) þ 33  > 11  :  3  9

9    50 22 79  22 79 > =     17 55 : 17 55  ¼  33 >   ;  3 8 9 8 9

Multiply the elements in the indicated row of the last determinant by 2 and add to the 2nd row to obtain   50    27   3

 79   37 :  8 9

22 1

Multiply the elements in the indicated row of the last determinant 22, 8 and add respectively to the elements in the 1st and 3rd rows to obtain   544    27   213

 0 735    544  1 37  ¼ (1) þ  213 0 287 

 735  ¼ 427: 287 

342

DETERMINANTS

[CHAP. 28

28.27 Factor the following determinant.  x   2 x   x3

y y2 y3

  1 1     1  ¼ xy x    x2 1

1 y y2

 1   1  1

Removing factors x and y from 1st and 2nd columns respectively :

  0   ¼ xy x  1   x2  1

 1   y  1 1  y2  1 1 

  x1 ¼ xy 2 x 1

 y  1  y2  1 

0

Adding 1 times elements in 3rd column to the corresponding elements in 1st and 2nd columns.

   1 1   ¼ xy(x  1)(y  1) x þ 1 y þ 1

Removing factors (x  1) and (y  1) from 1st and 2nd columns respectively:

¼ xy(x  1)(y  1)(y  x):

28.28 Solve the system 2x þ y  z þ w ¼ 4 x þ 2y þ 2z  3w ¼ 6 3x  y  z þ 2w ¼ 0 2x þ 3y þ z þ 4w ¼ 5: SOLUTION   2 1 1 1   1 2 2 3   D¼  ¼ 86  3 1 1 2   2 3 1 4   4   6  D1 ¼   0   5

1 2 1 3

 1 1  2 3   ¼ 86 1 2  1 4

  2 1 4 1   1 2 6 3   D3 ¼   ¼ 258  3 1 0 2   2 3 5 4 Then

x¼

D1 ¼ 1, D

y¼

  2 4  1 6  D2 ¼  3 0   2 5

1 2 1 1

 1  3   ¼ 172 2  4

  2 1 1 4    1 2 2 6   D4 ¼   ¼ 86  3 1 1 0   2 3 1 5  D2 ¼ 2, D

z¼

D3 ¼ 3, D

w¼

D4 ¼ 1: D

CHAP. 28]

343

DETERMINANTS

28.29 The currents i1 , i2 , i3 , i4 , i5 (measured in amperes) can be determined from the following set of equations. Find i3 . i1  2i2 þ i3 ¼ 3 i2 þ 3i4  i5 ¼ 5 i1 þ i2 þ i3  i5 ¼ 1 2i2 þ i3  2i4  2i5 ¼ 0 i1 þ i3 þ 2i4 þ i5 ¼ 3 SOLUTION    1 2 3 0 0   0 1 5 3 1   D3 ¼  1 1 1 0 1  ¼ 38, 0 2 0 2 2   1 0 3 2 1

29.30 Determine whether the system

 1  0  D ¼  1 0  1

2 1 1 2 0

 1 0 0  0 3 1  1 0 1  ¼ 19, 1 2 2  1 2 1

i3 ¼

D3 ¼ 2 amp. D

x  3y þ 2z ¼ 4 2x þ y  3z ¼ 2 4x  5y þ z ¼ 5

is consistent.    1 3 2   D ¼  2 1 3  ¼ 0:  4 5 1    4 3 2   D1 ¼  2 1 3  ¼ 7:  5 5 1

SOLUTION

However,

Hence at least one of the determinants D1 , D2 , D3 = 0 so that the equations are inconsistent. This could be seen in another way by multiplying the first equation by 2 and adding to the second equation to obtain 4x  5y þ z ¼ 6 which is not consistent with the last equation.

28.31 Determine whether the system 4x  2y þ 6z ¼ 8 2x  y þ 3z ¼ 5 2x  y þ 3z ¼ 4 is consistent. SOLUTION

 4   D ¼ 2  2  4   D2 ¼  2  2

 2 6   1 3  ¼ 0  1 3   8 6   5 3 ¼ 0  4 3

 8   D1 ¼  5  4  4   D3 ¼  2  2

 2 6   1 3  ¼ 0  1 3   2 8   1 5  ¼ 0  1 4 

Nothing can be said about the consistency from these facts. On closer examination of the system it is noticed that the second and third equations are inconsistent. Hence the system is inconsistent.

344

DETERMINANTS

[CHAP. 28

28.32 Determine whether the system 2x þ y  2z ¼ 4 x  2y þ z ¼ 2 5x  5y þ z ¼ 2 is consistent. SOLUTION D ¼ D1 ¼ D2 ¼ D3 ¼ 0. Hence nothing can be concluded from these facts. 3 4 Solving the first two equations for x and y (in terms of z), x ¼ (z þ 2), y ¼ (z þ 2). These values are found 5 5 by substitution to satisfy the third equation. (If they did not satisfy the third equation the system would be inconsistent.) 3 4 Hence the values x ¼ (z þ 2), y ¼ (z þ 2) satisfy the system and there are infinite sets of solutions, obtained 5 5 by assigning various values to z. Thus if z ¼ 3, then x ¼ 3, y ¼ 4; if z ¼ 2, then x ¼ 0, y ¼ 0; etc. It follows that the given equations are dependent. This may be seen in another way by multiplying the second equation by 3 and adding to the first equation to obtain 5x  5y þ z ¼ 2 which is the third equation.

28.33 Does the system 2x  3y þ 4z ¼ 0 x þ y  2z ¼ 0 3x þ 2y  3z ¼ 0 possess only the trivial solution x ¼ y ¼ z ¼ 0? SOLUTION

   2 3 4   D ¼  1 1 2  ¼ þ7 3 2 3 

D1 ¼ D2 ¼ D3 ¼ 0

Since D = 0 and D1 ¼ D2 ¼ D3 ¼ 0, the system has only the trivial solution.

28.34 Find nontrivial solutions for the system x þ 3y  2z ¼ 0 2x  4y þ z ¼ 0 xþyz¼0 if they exist. SOLUTION

  1 3 2   D ¼  2 4 1  ¼ 0 1 1 1 

D1 ¼ D2 ¼ D3 ¼ 0

Hence there are nontrivial solutions. To determine these nontrivial solutions solve for x and y (in terms of z) from the first two equations (this may not always be possible). We find x ¼ z/2, y ¼ z/2. These satisfy the third equation. An infinite number of solutions is obtained by assigning various values to z. For example, if z ¼ 6, then x ¼ 3, y ¼ 3; if z ¼ 4, then x ¼ 2, y ¼ 2; etc.

CHAP. 28]

345

DETERMINANTS

28.35 For what values of k will the system x þ 2y þ kz ¼ 0 2x þ ky þ 2z ¼ 0 3x þ y þ z ¼ 0 have nontrivial solutions? SOLUTION Nontrivial solutions are obtained when   1 2 k   D ¼  2 k 2  ¼ 0: 3 1 1 Hence D ¼ 3k2 þ 3k þ 6 ¼ 0 or k ¼ 1, 2.

Supplementary Problems 28.36

Evaluate   4 (a)  1   2 (b)  3

each of the following determinants.     2 1  3    (e) (c) 4 0 2     2x 3y  4   (f) (d )  4x y  7

  aþ b a b    a b     2x  1 x þ 1     x þ 2 x  2

28.37

Show that if the elements of one row (or column) of a second-order determinant are multiplied by the same number, the determinant is multiplied by the number.

28.38

Solve the unknowns in each of the following systems. (a)

(b)

28.39

28.40



5x þ 2y ¼ 4 2x  y ¼ 7

(c)

3r  5s ¼ 6 4r þ 2s ¼ 5

(d )



28 þ 4x þ 5y ¼ 0 3x þ 4y þ 10 ¼ 0

(e)

5x  4y ¼ 16 2x þ 3y ¼ 10

(f)

Evaluate each determinant.    2 1 2   (c) (a)  3 1 3   1 3 2     1 0 2   (d ) (b)  0 3 4   4 2 1  For what value of k does

   3 1 4    2 1 3    1 3 2     x y z    2 3 1     4 1 2

8 x3 yþ4 > > < 3 þ 5 ¼7 > > : x þ 2  y  6 ¼ 3 7 2 8 3x þ 2y þ 1 > ¼4 > < xþy > > : 5x þ 6y  7 ¼ 2 xþy

 1  (e)  a  a2

1 b b2

  k þ3 1 2    3 2 1  ¼ 0?   k 3 3

 1  c  c2 

( g)

84 1 2 > >

> :3 5 ¼  1 x y 12 8 4 3 > > < 3u  5v ¼ 1 (h) > > : 1  1 ¼ 1 u v 6

346

28.41

28.42

28.43

28.44

DETERMINANTS

[CHAP. 28

Show that if the elements of one row (or column) of a third-order determinant are multiplied by the same number, the determinant is multiplied by the number. Solve for the unknowns in each of the following systems. 8 8 < 3x þ y  2z ¼ 1 < u þ 2v  3w ¼ 7 (a) 2x þ 3y  z ¼ 2 (b) 2u  v þ w ¼ 5 : : x  2y þ 2z ¼ 10 3u  v þ 2w ¼ 8 Solve for the indicated unknown. 8 < 3i1 þ i2  2i3 ¼ 0 i þ 2i2  3i3 ¼ 5 for i2 (a) : 1 2i1  i2 þ i3 ¼ 1

8 < 2x þ 3y ¼ 2 (c) 5y  2z ¼ 4 : 3z þ 4x ¼ 7

8 < 1=x þ 2=y þ 1=z ¼ 1=2 (b) 4=x þ 2=y  3=z ¼ 2=3 : 3=x  4=y þ 4=z ¼ 1=3

for x

(a) Prove Property I: If the rows and columns of a determinant are interchanged, the value of the determinant is the same. (b) Prove Property II: If each element in a row (or column) is zero, the value of the determinant is zero.

28.45

Show that the determinant  1 2  2 4  3 8   4 16

3 6 12 24

 4  3  2  1

  2 4   1 2   3 1   4 3

1 2 3 2

 3  4  2  1 

equals zero. 28.46

Transform the determinant

into an equal determinant having three zeros in the 3rd column. 28.47

Without changing the value of the determinant    4 2 1 3 1    2 1 3 2 2    3 4 2 1 3     1 3 4 1 1    2 1 2 4 2 obtain four zeros in the 4th column.

28.48

For the determinant   1 2   4 1   3 1   2 4

3 2 2 1

 2  2  1  3

(a) write the minors and cofactors of the elements in the 3rd row, (b) express the value of the determinant in terms of minors or cofactors, (c) find the value of the determinant.

CHAP. 28]

28.49

347

DETERMINANTS

Transform the determinant    2 1 2 3    3 2 3 2     1 2 1 2    4 3 1 3  into a determinant having three zeros in a row and then evaluate the determinant by use of expansion by minors.

28.50

Evaluate each determinant.   1 2 1   2 3 2 (c)  1  3 1  1 4 3   3 2 1   2 0 3  (d)  1 3 2  2 4 1   1 1 2

   2 1 3 2    3 1 2 4  (a)   1 3 1 3    1 2 2 3     3 1 2 1    4 2 0 3  (b)  2 1 3 2    1 3 1 4 28.51

28.53

3 4 1 0 1

 2  3  0  1  0

Factor each determinant:  a  (a)  a2  a3

28.52

 1  1  4  2

b b2 b3

 c  c2  c3 

 1  1 (b)  1 1

Solve each system: 8 x  2y þ z  3w ¼ 4 > > < 2x þ 3y  z  2w ¼ 4 (a) 3x  4y þ 2z  4w ¼ 12 > > : 2x  y  3z þ 2w ¼ 2

1 x x2 x3

1 y y2 y3

 1  z  z2  z3 

8 2x þ y  3z ¼ 5 > > < 3y þ 4z þ w ¼ 5 (b) 2z  w  4x ¼ 0 > > : w þ 3x  y ¼ 4

Find i1 and i4 for the system 8 2i1  3i3  i4 ¼ 4 > > > > < 3i1 þ i2  2i3 þ 2i4 þ 2i5 ¼ 0 i1  3i2 þ 2i4 þ 3i5 ¼ 2 > > > i1 þ 2i3  i5 ¼ 9 > : 2i1 þ i2 ¼ 5

28.54

Determine whether each system is consistent. 8 < 2x  3y þ z ¼ 1 (a) x þ 2y  z ¼ 1 : 3x  y þ 2z ¼ 6

28.55

8 < 2x  y þ z ¼ 2 (b) 3x þ 2y þ 4z ¼ 1 : x  4y þ 6z ¼ 3

8 < x þ 3y  2z ¼ 2 (c) 3x  y  z ¼ 1 : 2x þ 6y  4z ¼ 3

Find non-trivial solutions, if they exist, for the system 8 < 3x  2y þ 4z ¼ 0 2x þ y  3z ¼ 0 : x þ 3y  2z ¼ 0

8 < 2u þ v  3w ¼ 1 (d) u  2v  w ¼ 2 : u þ 3v  2w ¼ 2

348

28.56

DETERMINANTS

[CHAP. 28

For what value of k will the system 8 2x þ ky þ z þ w ¼ 0 > > < 3x þ (k  1)y  2z  w ¼ 0 x  2y þ 4z þ 2w ¼ 0 > > : 2x þ y þ z þ 2w ¼ 0 possess non-trivial solutions?

ANSWERS TO SUPPLEMENTARY PROBLEMS (b) 2

28.36

(a) 5

(c) 4

28.38

(a) (b) (c) (d )

28.39

(a) 43

28.40

All values of k.

28.42

(a) x ¼ 2, y ¼ 1, z ¼ 3; (2, 1, 3) (b) u ¼ 1, v ¼ 1, w ¼ 2; (1, 1, 2) (c) x ¼ 4, y ¼ 2, z ¼ 3; (4, 2, 3)

28.43

(a) i2 ¼ 0:8

28.48

(c) 38

28.49

28

28.50

(a) 38

28.51

(a) abc(a  b)(b  c)(c  a)

28.52

(a) x ¼ 2, y ¼ 1, z ¼ 3, w ¼ 1

28.53

i1 ¼ 3, i4 ¼ 2

28.54

(a) consistent

28.55

Only the trivial solution x ¼ y ¼ z ¼ 0.

28.56

k ¼ 1

x ¼ 2, y ¼ 3; (2, 3) r ¼ 1=2, s ¼ 3=2; (1=2, 3=2) x ¼ 2, y ¼ 4; (2, 4) x ¼ 8=23, y ¼ 82=23; (8=23, 82=23) (b) 19

(e) a2  b2

(d ) 14xy

(e) (f ) (g) (h)

x ¼ 12, y ¼ 16; (12, 16) x ¼ 5, y ¼ 2; (5, 2) x ¼ 12, y ¼ 15; (12, 15) u ¼ 2=3, v ¼ 3=5; (2=3, 3=5)

(d ) 5x þ 8y  14z

(c) 0

( f ) x2  8x

(e) bc2  cb2 þ a2 c  ac2 þ ab2  ba2

(b) x ¼ 6

(b) 143

(c) 108

(d ) 88

(b) (x  1)(y  1)(z  1)(x  y)(y  z)(z  x)

(b) dependent

(b) x ¼ 1, y ¼ 1, z ¼ 2, w ¼ 0

(c) inconsistent

(d) inconsistent

CHAPTER 29

Matrices 29.1

DEFINITION OF A MATRIX

A matrix is a rectangular array of numbers. The numbers are the entries or elements of the matrix. The following are examples of matrices.

1 7

2

 4 , 0

3 2 7 5, 3

1 4 4 1

2

3 5 4 3 5, 8



8 1 5 3

0 6



Matrices are classified by the number of rows and columns. The matrices above are 2  2, 3  2, 3  1, and 2  3 with the first number indicating the number of rows and the second number indicating the number of columns. When a matrix has the same number of rows as columns, it is a square matrix. 2

a12 a22 .. . am2

a11 6 a21 6 A¼6 . 4 .. am1

a13 a23 .. . am3

3 a1n a2n 7 7 .. 7 . 5 . . . amn ... ...

The matrix A is an m  n matrix. The entries in matrix A are double subscripted with the first number indicating the row of the entry and the second number indicating the column of the entry. The general entry for the matrix is denoted by aij . The matrix A can be denoted by [aij ].

29.2

OPERATIONS WITH MATRICES

If matrix A and matrix B have the same size, same number of rows and same number of columns, and the general entries are of the form aij and bij respectively, then the sum A þ B ¼ ½aij  þ ½bij  ¼ ½aij þ bij  ¼ ½cij  ¼ C for all i and j.

EXAMPLE 29.1.

Find the sum of A ¼

AþB¼

2 3 6 0

2 6



4 0 þ 1 1

3 0

4 1 3 1



and

B¼



2 2þ0 ¼ 2 6 þ (1)

349

0 3 1 1

3þ3 0þ1

 2 : 2



 4 þ (2) 2 6 2 ¼ 1 þ 2 5 1 1

350

MATRICES

[CHAP. 29

The matrix 2A is called the opposite of matrix A and each entry in 2A is the opposite of the corresponding entry in A.

A¼

Thus, for



 2 3 , 0 1

1 2

A ¼

2 0

1 2

3 1



Multiplying a matrix by a scalar (real number) results in every entry in the matrix being multiplied by the scalar.

EXAMPLE 29.2.

2 3 Multiply the matrix A ¼ 6 0

4 1

2 3 2A ¼ 2 6 0

 by 22. 

4 4 ¼ 1 12

6 0

8 2



The product AB where A is an m  p matrix and B is a p  n matrix is C, an m  n matrix. The entries cij in matrix C are found by the formula cij ¼ ai1 b1j þ ai2 b2j þ ai3 b3j þ    þ aip bpj . 2

a11 6 a21 6 6 . 6 .. 6 6 6 ai1 6 6 .. 4 . am1

A a13 a23 .. . ai3 .. . am3

a12 a22 .. . ai2 .. . am2

  .. .  .. . 

3 a1p 2 b11 a2p 7 7 7 6 .. 7 6 b21 . 7 6 b31 76 aip 7 6 . 7 4 .. .. 7 . 5 bp1 amp

b12 b22 b32 .. . bp2

   .. . 

AB ¼

AB ¼

AB ¼

AB ¼

2 4 Find the product AB when A ¼ 0 1 

2

3 0 1 6 1 3 4 2 4 0

2 4 0 1

1 1 3

   .. . 

b1j b2j b3j .. . bpj



EXAMPLE 29.3.

¼2

B

1 2

c11 3 6 c21 b1n 6 b2n 7 6 . 7 6 .. b3n 7 ¼ 6 7 6 6 ci1 .. 7 . 5 6 6 .. 4 . bpn cm1

c12 c22 .. . ci2 .. . cm2

  .. .  .. . 

2

 and

3 0 B ¼ 4 1 3 4 0

C c1j c2j .. . cij .. . cmj

  .. .  .. . 

3 1 2 5: 2

1 1 3

3 1 7 25 2

 2(3) þ 4(1) þ 1(4) 2(0) þ 4(3) þ 1(0) 2(1) þ 4(1) þ 1(3) 2(1) þ 4(2) þ 1(2) 0(3) þ 1(1) þ (2)(4) 0(0) þ 1(3) þ (2)(0) 0(1) þ 1(1) þ (2)(3) 0(1) þ 1(2) þ (2)(2) 6  4 þ 4 0 þ 12 þ 0 2 þ 4 þ 3 018 0þ 3þ0 0þ16 6 9

12 9 4 3 5 6

2 þ 8  2 0þ2þ4







EXAMPLE 29.4.

1 2 3 Find the products CD and DC when C ¼ 1 0 4

2  1 1 2 3 6 CD ¼ 40 1 0 4 4

3

2

 and

1 D ¼ 40 4

3

1(1) þ 2(0) þ 3(4) 7 25 ¼ 1(1) þ 0(0) þ 4(4) 2

1(3) þ 2(2) þ 3(2) 1(3) þ 0(2) þ 4(2)



3 3 2 5: 2

3 c1n c2n 7 7 .. 7 . 7 7 7 cin 7 7 .. 7 . 5 cmn

CHAP. 29]

351

MATRICES

CD ¼

1 þ 0 þ 12 1 þ 0 þ 16

2



 3 þ 4  6 13 5 ¼ 3þ08 15 5

3 1 3

6 7 1 2 DC ¼ 4 0 25 1 0 4 2 2

1þ3 6 DC ¼ 4 0  2 4þ2

2

1(1) þ (3)(1) 6 ¼ 4 0(1) þ 2(1) 4 4(1) þ (2)(1) 3



3 2 4 2 2 þ 0 3  12 7 6 0 þ 0 0 þ 8 5 ¼ 4 2 0 8 þ 0 12  8 6 8

3 1(2) þ (3)(0) 1(3) þ (3)(4) 7 0(2) þ 2(0) 0(3) þ 2(4) 5 4(2) þ (2)(0) 4(3) þ (2)(4)

3 9 7 85 4

In Example 29.4, note that although both products CD and DC exist, CD = DC. Thus, multiplication of matrices is not commutative. An identity matrix is an n  n matrix with entries of 1 when the row and column numbers are equal and 0 everywhere else. We denote the n  n identity matrix by I n. For example, 2 3

 1 0 0 1 0 I2 ¼ and I3 ¼ 4 0 1 0 5: 0 1 0 0 1 If A is a square matrix and I is the identity matrix the same size as A, then AI ¼ IA ¼ A. 







 2 3 2 3 1 0 1 0 2 3 ¼ and AI ¼ , we use I ¼ For A ¼ 7 9 7 9 0 1 0 1 7 9





 1 0 2 3 2 3 and IA ¼ ¼ : 0 1 7 9 7 9

29.3

ELEMENTARY ROW OPERATIONS

Two matrices are said to be row equivalent if one can be obtained from the other by a sequence of elementary row operations.

Elementary Row Operations (1) Interchange two rows. (2) Multiplying a row by a nonzero constant. (3) Add a multiple of a row to another row. A matrix is said to be in reduced row-echelon form if it has the following properties: (1) All rows consisting of all zeros occur at the bottom of the matrix. (2) A row that is not all zeros has a 1 as its first non-zero entry, which is called the leading 1. (3) For two successive non-zero rows, the leading 1 in the higher row is further to the left than the leading 1 in the lower row. (4) Every column that contains a leading 1 has zeros in every other position in the column.

352

MATRICES

EXAMPLE 29.5.

[CHAP. 29

Use elementary row operations to put the matrix A in reduced row-echelon form when

2

3 2 1 4 A ¼ 41 3 2 5: 3 1 6  is the symbol used between two matrices to indicate that the two matrices are row equivalent. R2 in front of a matrix means that the row following it was row 2 in the previous matrix. R3  3R1 in front of a matrix means that the row following it was obtained from the previous matrix by subtracting 3 times row 1 from row 3. 2 3 R2 1 2 1 4 6 6 7 A ¼ 41 3 2 5  R1 4 2 3 3 1 6 2

3 3 3 2 2 3 2 1 3 2 1 3 2 7 7 7 6 6 1 4 5  R2  2R1 4 0 5 0 5  15 R2 4 0 1 05 1 6 0 10 0 R3  3R1 0 10 0

R1  3R2

2

1 0 6  40 1 R3 þ 10R2 0 0

3 2 7 05 0 2

3 1 0 2 The reduced row-echelon form of matrix A is ¼ 4 0 1 0 5: 0 0 0

29.4

INVERSE OF A MATRIX

A square matrix A has an inverse if there is a matrix A1 such that AA1 ¼ A1 A ¼ I. To find the inverse, if it exists, of a square matrix A we complete the following procedure. (1) Form the partitioned matrix [AjI], where A is the given n  n matrix and I is the n  n identity matrix. (2) Perform elementary row operations on [AjI] until the partitioned matrix has the form [IjB], that is, until the matrix A on the left is transformed into the identity matrix. If A cannot be transformed into the identity matrix, matrix A does not have an inverse. (3) The matrix B is A1 , the inverse of matrix A. 2

EXAMPLE 29.6.

3 2 5 4 Find the inverse of matrix A ¼ 4 1 4 3 5: 1 3 2

 4  1  6 ½AjI ¼6 4 3  0 41  1 3 2  0 2 1 6  R2  2R1 6 40 R3  R1 0 2 R1  4R2 1 6 60  4 2

2

5

R3 þ 7R2

0 0

3

R2

2

1

7 6 6 1 07 5  R1 4 2 0 1 1  1 4 3  0  3 2  1 2  7 5  0 1  0 1=3  4=3  1 2=3  1=3  0 0 1=3  7=3

 3 3  0 1 0  7 5 4  1 0 0 7 5  3 2  0 0 1  3 2 0 0 1 1 4 3   7 6 1 6 07 1 2=3  1=3 2=3 5   3 R2 4 0  1 0 1 0 7 5   3 2 5=3 0 1 0 1=3  4=3  7 6 6 0 1 2=3  1=3 2=3 0 7  5 4  3R3 0 0 11=3 1 1 7 4

0

3

7 07 5

1 5=3 2=3

0

3

7 07 5

11 3

CHAP. 29]

 3 1 0 0  1 2 1  6 7 8 2 5 ¼ ½IjA1   R2  (2=3)R3 4 0 1 0  5   0 0 1 7 11 3 2 3 1 2 1 6 7 ¼ 4 5 8 25 7 11 3 R1  (1=3)R3

A1

353

MATRICES

2

If the matrix A is row equivalent to I, then the matrix A has an inverse and is said to be invertible. A does not have an inverse if it is not row equivalent to I. 2

1 EXAMPLE 29.7. Find the inverse, if it exists, for matrix A ¼ 4 2 1   3 2 2 1 3 4  1 0 1 3 4  1 0 0   7 6 6   6 7 ½AjI  ¼6 4 2 5 3  0 1 0 5  R2 þ 2R1 4 0 1 5  2 1   R3  R1 0 1 5  1 0 1 4 90 0 1

0

3 5 4 3

3 4 3 5: 9

7 07 5 1

 3 1 0 11  5 3 0  7 6 60 1  5  2 1 07 5 4  R3  R2 0 0 0  3 1 1 R1  3R2

2

The matrix A is row equivalent to the matrix on the left. Since the matrix on the left has a row of all zeros, it is not row equivalent to I. Thus, the matrix A does not have an inverse.

Another way to determine whether the inverse of a matrix A exists is that the determinant associated with an invertible matrix is non-zero, that is, det A = 0 if A1 exists. For 2  2 matrices, the inverse can be found by a special procedure:

a If A ¼ 11 a21

a12 a22

 then

1

A

1 a22 ¼ det A a21

a12 a11

 where det A = 0:

(1) Find the value of det A. If det A = 0, then the inverse exists. (2) Exchange the entries on the main diagonal, swap a11 and a22 . (3) Change the signs of the entries on the off diagonal, replace a21 by a21 and a12 by a12 . (4) Multiply the new matrix by 1/det A. This product is A1 . 29.5

MATRIX EQUATIONS

A matrix equation AX ¼ B has a solution if and only if the matrix A1 exists and the solution is X ¼ A1 B.

   x 12 EXAMPLE 29.8. Solve the matrix equation ¼ : y 6





 1 3 5 3=11 5=11 7 5 or If A ¼ then A1 ¼ 2=11 7=11 2 3 11 2 7 7 5 2 3

354

MATRICES

[CHAP. 29



 

  1 3 5 7 5 x 1 3 5 12 ¼ 11 2 7 2 3 y 11 2 7 6

 

 0 x 1 11 1 6 ¼ 11 11 18 0 11 y



  1 0 x 1 6 ¼ 11 18 0 1 y



 x 6=11 ¼ y 18=11

29.6

MATRIX SOLUTION OF A SYSTEM OF EQUATIONS

To solve a system of equations using matrices, we write a partitioned matrix which is the coefficient matrix on the left augmented by the constants matrix on the right. x þ 2y þ 3z ¼ 6 The augmented matrix associated with the system x  z ¼ 0 is x  y  z ¼ 4 2

1 A ¼ 41 1 EXAMPLE 29.9.

2 0 1

 3 3  6 1  0 5 1  4

Use matrices to solve the system of equations: x2 þ x3  2x4 ¼ 3 ¼ 2 x1 þ 2x2  x3 2x1 þ 4x2 þ x3  3x4 ¼ 2 x1  4x2  7x3  x4 ¼ 19

Write the augmented matrix for the system.

 3 0 1 1 2  3 61 2 1 0  0 7 7 6 42 4 1 3  2 5  1 4 7 1 19 2

Put the matrix on the left in reduced row-echelon form. 2 R2 1  R1 6 60 6 42 1

3 2  2 1 0 2 1  60 1 1 2  3 7 7 6  7 6 4 1 3  2 5 R3  2R1 4 0 4 7 1  19 0 R4  R1

R1  2R2  (1=3)R3 R4 þ 6R2

2

1 0 60 1 6 6 40 0 0 0

2 R1  R4 1 0  R2 þ R4 6 60 1 6 R3 þ R4 4 0 0 0 0

3  2 1 0 2  1 1 2  3 7 7  7 0 3 3  6 5 6 6 1  21

3 2  4 8 R1 þ 3R3 1  6 2  3 7 7  R2  R3 6 0  7 6 40 1  2 5 (1=3)R4 0 0 13  39

3 1 1

3  0 0  1  0 0  2 7 7  7 1 0  1 5 0 1 3

3  0 0 1 2  1 0 1  1 7 7  7 0 1 1  2 5 0 0 1 3

CHAP. 29]

355

MATRICES

From the reduced row-echelon form of the augmented matrix, we write the equations: x1 ¼ 1, x2 ¼ 2, x3 ¼ 1, and x4 ¼ 3: Thus, the solution of the system is (1, 2, 1, 3). x1 þ 2x2  x3 ¼ 0 ¼1 3x1 þ 5x2  #   # # " " " 1 2 1  0 1 2 1  0 1 2 1  0      R2 0 1 3  1 R2  3R1 0 1 31 3 5 01

EXAMPLE 29.10. Solve the system of equations:

 R1  2R2

x1 þ 5x3 ¼ 2

"

 # 5  2  0 1 3  1 1 0

x2  3x3 ¼ 1

and

x1 ¼ 2  5x3

Thus,

x2 ¼ 1 þ 3x3 .

and

The system has infinitely many solutions of the form (2  5x3 , 1 þ 3x3 , x3 ), where x3 is a real number.

Solved Problems 29.1

Find (a) A þ B, (b) A 2 B, (c) 3A, and (d ) 5A 2 2B when " A¼

2

1 1

1

1 4

#

22

¼ " (c) 3A ¼ 3

1  (3)

1  (3) 2

1

1

1 1

4

1  1

#

" ¼

B¼

and

SOLUTION



2 1 1 2 3 (a) A þ B ¼ þ 1 1 4 3 1

2þ2 1 þ (3) ¼ 1 þ (3) 1 þ 1



2 1 1 2 3 (b) A  B ¼  1 1 4 3 1 "

"

4 2

2

3

4

3

1

2





 1þ4 4 2 5 ¼ 4 þ (2) 4 0 2  4 2 #

14 4  (2)

" ¼

3(2)

3(1)

3(1)

3(1)

3(1)

3(4)

4 3

0

2 2

#

" ¼

6

#

6 3

3 3 12



 2 1 1 2 3 4 (d ) 5A  2B ¼ 5 2 1 1 4 3 1 2 " # 5(2)  2(2) 5(1)  2(3) 5(1)  2(4) ¼ 5(1)  2(3) 5(1)  2(1) 5(4)  2(2) " ¼

6

11 3

1 7

24

#

3

#

#

356

MATRICES

29.2

Find, if they exist, (a) AB, (b) BA, and (c) A 2 when

A¼ 3

2

1

[CHAP. 29

2 3 2 B ¼ 435 0

and

SOLUTION

2 3 2 (a) AB ¼ ½ 3 2 1 4 3 5 ¼ ½3(2) þ 2(3) þ 1(0) ¼ ½12 0 2 3 2 3 2 2(3) 2(2) 2(1) 6 4 2 (b) BA ¼ 4 3 5½ 3 2 1  ¼ 4 3(3) 3(2) 3(1) 5 ¼ 4 9 6 0 0 0 0(3) 0(2) 0(1) (c) A2 ¼ ½ 3

29.3

3 2 35 0

2 1  ½ 3 2 1 ; not possible. An , n . 1, exists for square matrices only.

Find AB, if possible. 2 3 2 1 (a) A ¼ 4 3 4 5 1 6 2 3 1 3 (b) A ¼ 4 4 5 5 0 2

2

and

0 B ¼ 44 8

and

1 B¼ 0

1 0 1 2 7

3 0 25 7



SOLUTION 2

32 3 2 1 0 1 0 (a) AB ¼ 4 3 4 54 4 0 2 5; 1 6 8 1 7

not possible.

A is a 3  2 matrix and B is a 3  3. Since A has only two columns it can only multiply matrices having two rows, 2  k matrices. 2

29.4

3 2  1 3

1(1) þ 3(0) 1 2 (b) AB ¼ 4 4 5 5 ¼ 4 4(1) þ (5)(0) 0 7 0 2 0(1) þ 2(0)

3 2 3 1(2) þ 3(7) 1 19 4(2) þ (5)(7) 5 ¼ 4 4 27 5 0(2) þ 2(7) 0 14

Write each matrix in reduced row-echelon form. 2 2 3 1 2 1 1 0 1 3 6 2 3 2 3 (a) 4 2 3 1 5 (b) 6 4 3 5 3 4 4 5 2 1 1 1 2

3 4 1 7 7 35 5

SOLUTION 2 3 2 0 1 3 R2 2 (a) 4 2 3 1 5  R1 4 0 4 5 2 4

3 2 3 3 1 2 3 1 40 1 3 5  1 3 5  5 2 0 R3  2R1 0 1 2 3 2 4 R1  4R3 1 0 (1=2)R1 1 0 4 0 1 3 5  R2 þ 3R3 4 0 1  1 0 0 (1=3)R3 0 0

3 2 0 8 4 0 1 3 5 R3 þ R2 0 0 3 3 0 05 1

R1  3R2

2

CHAP. 29]

357

MATRICES

2

0 The reduced row-echelon form of 4 2 4

3 1 3 3 1 5 5 2

2 is

3 1 0 0 4 0 1 0 5: 0 0 0

2

3 2 3 1 2 1 1 4 1 2 1 1 4 6 2 3 6 2 3 1 7 1 0 1 9 7 7  R2  2R1 6 0 7 (b) 6 4 3 5 3 4 3 5 R3  3R1 4 0 1 0 1 9 5 1 1 1 2 5 R4 þ R1 0 1 0 1 9 2 3 R1 þ 2R2 1 0 1 3 14 6 0 1 0 1 9 7 6 7  R3  R2 4 0 0 0 0 05 0 0 R4 þ R2 0 0 0 2

1 2 6 2 3 6 The reduced row-echelon form of 4 3 5 1 1

29.5

1 2 3 1

Find the inverse, if it exists, for each matrix.



 3 6 2 3 (b) B ¼ (a) A ¼ 1 2 1 7

3 1 4 3 1 7 7 4 35 2 5

2

1 (c) C ¼ 4 1 6

2 is

1 60 6 40 0

1 0 2

3 0 1 5 3

0 1 0 0

3 1 3 14 0 1 9 7 7 0 0 05 0 0 0

2

3 (d) D ¼ 4 1 0

2 0 1

3 1 1 5 2

SOLUTION   2 3  ¼ 14  3 ¼ 17; det A = 0 so A1 exists: (a) det A ¼  1 7 



 1 7 3 7=17 3=17 A1 ¼ or A1 ¼ 1=17 2=17 2 17 1 The first form of the matrix is frequently used because it reduces the amount of computation with fractions that needs to be done. Also, it makes it easier to work with matrices on a graphing calculator.    3 6   ¼ 6  6 ¼ 0: Since det B ¼ 0, B 21 does not exist. (b) det B ¼  1 2   3 3 2 2 1 1 0  1 0 0 1 1 0  1 0 0 1 1  1 1 0 5 (c) ½CjI  ¼ 4 1 0 1  0 1 0 5  4 0 0 4 3  6 0 1 6 2 3  0 0 1   3 2 2 3 1 0 0  2 3 1 1 0 1 0 1  0  4 0 1 1  1 1 0 5  4 0 1 0  3 3 1 5 ¼ ½IjC1   0 0 1  2 4 1 0 0 1 2 4 1 2 3 2 3 1 C1 ¼ 4 3 3 1 5 2 4 1    3 2 2 3 2 3 3 2 1  1 0 0 1 0 1  0 1 0 1 0 1  0 1 0 (d ) ½DjI  ¼ 4 1 0 1  0 1 0 5  4 3 2 1  1 0 0 5  4 0 2 4  1 3 0 5   0 1 2 0 0 1 0 1 2 0 0 1 0 1 0 1 20   3 3 2 2 1 0 1 0 1  0 1 0 1  0 1 0 0 15 2  0  40 1 2  0 0 15  40 1 0 0 0  1 3 2 0 2 4  1 3 0 Since the left matrix in the last form is not row equivalent to the identity matrix I, D does not have an inverse.

358

MATRICES

2

29.6

3 2 1 A¼4 1 05 3 4

If

(a) 2X þ 3A ¼ B

[CHAP. 29

2

and

3 0 3 B¼4 2 0 5, solve each equation for X. 4 1

(b) 3A þ 6B ¼ 3X

SOLUTION (a) 2X þ 3A ¼ B: So 2X ¼ 3A þ B and X ¼  32 A þ 12 B. 2 3 2 3 2 2 1 0 3 3þ0 34 1 X¼ 1 05 þ 4 2 0 5 ¼ 4 (3=2) þ 1 2 2 3 4 4 1 (9=2)  2

3 2 3 (3=2) þ (3=2) 3 3 5 ¼ 4 1=2 0þ0 05 6 þ (1=2) 13=2 11=2

(b) 3A þ 6B ¼ 3X: So 3X ¼ 3A þ 6B and X ¼ A  2B. 2

2 X ¼ 4 1 3

29.7

3 3 2 3 2 3 2 2 5 2þ0 16 0 3 1 05 0 5 ¼ 4 1  4 0 þ 0 5 ¼ 4 5 0 5  24 2 5 6 4 1 3 þ 8 4 þ 2 4

Write the matrix equation AX ¼ B and use it to solve the system

x þ y ¼ 4 . 2x þ y ¼ 0

SOLUTION  X ¼ B    1 1 x 4 ¼ 2 1 y 0 A



1 1 2 1



1 1 2 1

A¼

1 1 2 1

 so

A1 ¼

1 1 1 2



 1 1 1 ¼ 2 1 1

 

  x 1 1 4 ¼ y 2 1 0

  x 4 ¼ y 8 The solution to the system is (4, 8).

29.8

Solve each system of equations using matrices. (a)

x  2y þ 3z ¼ 9 x þ 3y ¼ 4 2x  5y þ 5z ¼ 17

(b) x þ 2y  z ¼ 3 3x þ y ¼4 2x  y þ z ¼ 2

SOLUTION  3 2 2 1 1 2 3  9 (a) 4 1 3 0  4 5  4 0 0 2 5 5  17 2 1  40 0

  3 2  3 3 2 2 3  9 1 0 9  19 1 0 9  19 1 3  5 5  4 0 1 3  5 5  4 0 1 3  5 5 1 1  1 0 0 1 2 0 0 2 4  3  0 0 1 1 0  1 5 0 1 2

From the reduced row echelon form of the matrix, we write the equations: x ¼ 1, y ¼ 1, and z ¼ 2:

CHAP. 29]

359

MATRICES

The system has the solution (1, 1, 2). 2

1 (b) 4 3 2

 3 2 1 2 1  3 0  4 5  4 0 5 0 5 12

2 1 1

  3 3 2 1 2 1  3 1  3 3  5 5 3  5 5  4 0 5 0 0 0 1 3  4

Since the last row results in the equation 0z ¼ 1, which has no solution, the system of equations has no solutions.

29.9

A¼

2 5 0 7



B¼

Supplementary Problems 3 1=2 5 1 1 3





C¼

2 5=2 0 0 2 3





D¼ 7 3

Perform the indicated operations, if possible. (a) (b) (c) (d ) 29.10

BþC 5A 2C 2 6B 26B

(e) ( f) (g) (h)

0 (a) A ¼ 4 4 8 2 1 (b) A ¼ 4 2 3 2 1 (c) A ¼ 4 0 3 2

3 1 0 0 25 1 7

3 1 7 1 85 1 1 3 2 3 5 45 2 1 3

6 6 2 7 7 (d ) A ¼ 6 4 15 6

C 2 5A BC (DA)B A2

(m) (n) (o) ( p)

B2 D(AB) A3 DB þ DC

2

and

and

and

and

3 2 1 B ¼ 4 3 4 5 1 6 2 3 1 1 2 B ¼ 42 1 15 1 3 2 2 3 4 6 3 B¼4 5 4 45 1 0 1

B ¼ 10 12

Solve each system of equations using a matrix equation of the form AX ¼ B. (a)

29.12

(i) ( j) (k) (l )

Find the product AB, if possible. 2

29.11

3B þ 2C DA AD C2B

x y¼ 0 5x  3y ¼ 10

(b) x þ 2y ¼ 1 5x  4y ¼ 23

(c) 1:5x þ 0:8y ¼ 2:3 0:3x  0:2y ¼ 0:1

Write each matrix in reduced row echelon form. 2 3 2 3 2 5 3 3 2 1 3 1 (d ) 4 3 2 4 95 (a) 4 1 0 2 1 5 3 1 1 2 5 3 2 4 3 2 3 2 1 2 0 1 1 1 0 2 4 0 6 1 3 1 61 1 1 5 17 2 37 7 7 (e) 6 (b) 6 4 1 1 3 41 2 0 6 35 1 15 2 3 7 3 4 1 1 1 5 0 2 3 2 3 4 1 2 2 1 3 1 1 6 1 6 1 2 1 7 0 2 1 3 7 7 7 (c) 6 (f ) 6 4 3 4 1 0 45 2 1 4 35 1 0 2 3 2 2 3 1

(d ) 2x þ 3y ¼ 40 3x  2y ¼ 8

360

MATRICES

29.13

Find the inverse, if it exists, of each matrix. 2 3

 5 3 4 9 13 (a) (e) 4 3 2 5 5 2 3 7 4 6 2 3

 1 2 1 3 2 (b) (f ) 42 3 25 1 2 4 2 3 2 3 2 3 2 0 1 1 3 2 4 6 1 1 0 7 2 7 (c) 6 (g) 4 5 3 35 4 0 1 2 15 2 5 2 2 1 3 0 2 3 2 3 1 2 0 1 1 3 2 6 0 1 2 1 7 7 (h) 6 (d ) 4 2 4 15 4 2 3 1 35 5 1 3 1 3 2 0

29.14

Solve each system of equations using matrices. x3 ¼ 1 x1 þ 5x2 þ 3x2 ¼4 3x2  4x3 ¼ 4

(a)

x  2y þ 3z ¼ 1 x þ 3y ¼ 10 2x  5y þ 5z ¼ 7

(e)

(b)

x  3y þ z ¼ 1 2x  y  2z ¼ 2 x þ 2y  3z ¼ 1

( f ) 4x1 þ 3x2 þ 17x3 ¼ 0 5x1 þ 4x2 þ 22x3 ¼ 0 4x1 þ 2x2 þ 19x3 ¼ 0

(c)

x þ y  3z ¼ 1 y z¼ 0 x þ 2y ¼ 1

(g)

x1 þ x2 þ x3 þ x4 ¼ 6  x4 ¼ 0 2x1 þ 3x2 3x1 þ 4x2 þ x3 þ 2x4 ¼ 4 x1 þ 2x2  x3 þ x4 ¼ 0

(d ) 4x  y þ 5z ¼ 11 x þ 2y  z ¼ 5 5x  8y þ 13z ¼ 7

(h)

3x1  2x2  6x3 ¼ 4 3x1 þ 2x2 þ 6x3 ¼ 1 x1  x2  5x3 ¼ 3

ANSWERS TO SUPPLEMENTARY PROBLEMS

 5 2 5 29.9 (a) (i) not possible 1 1 0

 10 25 (b) ( j) not possible 0 35

 14 8 30 (c) (k) ½ 28 21 28  6 10 24



 18 3 30 4 45 (d ) (l ) 6 6 18 0 49

 13 7=2 15 (m) not possible (e) 3 1 3

( f ) 14 14 (n) ½ 28 21 28 

 8 335 (g) not possible (o) 0 343

 1 3 5 (h) (p) ½ 38 11 35  1 3 6

[CHAP. 29

CHAP. 29]

2

3

29.10

3 4 (a) 4 10 16 5 26 46

29.11

(a) (5, 5)

2

(b) (23, 2)

3 1 0 0 1 (a) 4 0 1 0 05 0 0 1 1 2

1 60 (c) 6 40 0 2

1 60 (e) 6 40 0

29.13

(a)

0 1 0 0

3

3 0 07 7 15 0

(c) (1, 1) 2 1 60 6 (b) 4 0 0

3 0 2 0 1 1 1 0 27 7 0 0 1 45 0 0 0 0

3 13 2 9



 1 2 2 4 1 3 2 3 7 6 5 1 7 1 6 6 5 12 19 11 7 (c) 3 35 18 4 3 0 1 12 7 5 2 3 13 7 11 1 4 1 7 35 (d ) 28 22 14 10 (b)

29.14

2

21 15 23 19 5 7 5

6 (b) 4 8 4

2

29.12

361

MATRICES

11 2 (c) 4 21 20 1 26 3 0 2 4 0 1 1 1 0 7 7 0 0 0 15 0 0 0 0

1 0 0 (d) 4 0 1 0 0 0 1 2

1 60 (f ) 6 40 0

0 1 0 0

(b) no solution (c) (2z  1, z, z) where z is a real number (d ) (z þ 3, z þ 1, z) where z is a real number ( f ) (0, 0, 0) (g) (1, 0, 3, 2) (h) no solution

0 0 1 0

3 1 1 5 2 3 1=5 1 12=5 37 7 3=5 25 0 0

2 3 8 2 7 1 4 53 2 37 5 (e) 15 26 1 19 2 3 13 4 7 1 4 (f ) 2 7 4 5 33 16 10 1 2 3 21 16 18 1 4 (g) 16 14 11 5 19 19 19 19 3 2 21 9 4 7 16 3 3 0 37 7 (h) 6 0 2 2 5 6 4 6 9 3 4 1

(a) (1, 3, 2)

(e) (4, 8, 5)

14 24 5 2

(d ) (8, 8)

2

3

2

3 60 72 6 20 24 7 7 (d ) 6 4 10 12 5 60 72

CHAPTER 30

Mathematical Induction 30.1

PRINCIPLE OF MATHEMATICAL INDUCTION

Some statements are defined on the set of positive integers. To establish the truth of such a statement, we could prove it for each positive integer of interest separately. However, since there are infinitely many positive integers, this case-by-case procedure can never prove that the statement is always true. A procedure called mathematical induction can be used to establish the truth of the state for all positive integers. Principle of Mathematical Induction Let P(n) be a statement that is either true or false for each positive integer n. If the following two conditions are satisfied: (1)

P(1) is true,

and

(2)

Whenever for n ¼ k, P(k) is true implies P(k þ 1) is true.

Then P(n) is true for all positive integers n. 30.2

PROOF BY MATHEMATICAL INDUCTION

To prove a theorem or formula by mathematical induction there are two distinct steps in the proof. (1) (2)

Show by actual substitution that the proposed theorem or formula is true for some one positive integer n, as n ¼ 1, or n ¼ 2, etc. Assume that the theorem or formula is true for n ¼ k. Then prove that it is true for n ¼ k þ 1.

Once steps (1) and (2) have been completed, then you can conclude the theorem or formula is true for all positive integers greater than or equal to a, the positive integer from step (1).

Solved Problems 30.1

Prove by mathematical induction that, for all positive integers n, 1þ2þ3þþn¼ 362

n(n þ 1) : 2

CHAP. 30]

363

MATHEMATICAL INDUCTION

SOLUTION Step 1.

The formula is true for n ¼ 1, since 1¼

Step 2.

1(1 þ 1) ¼ 1: 2

Assume that the formula is true for n ¼ k. Then, adding (k þ 1) to both sides, 1 þ 2 þ 3 þ    þ k þ (k þ 1) ¼

k(k þ 1) (k þ 1)(k þ 2) þ (k þ 1) ¼ 2 2

which is the value of n(n þ 1)/2 when (k þ 1) is substituted for n. Hence if the formula is true for n ¼ k, we have proved it to be true for n ¼ k þ 1. But the formula holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2. Then, since it holds for n ¼ 2, it holds for n ¼ 2 þ 1 ¼ 3, and so on. Thus the formula is true for all positive integers n.

30.2

Prove by mathematical induction that the sum of n terms of an arithmetic sequence a, a þ d,   n a þ 2d,    is ½2a þ (n  1)d , that is 2 n a þ (a þ d ) þ (a þ 2d ) þ    þ ½a þ (n  1)d  ¼ ½2a þ (n  1)d : 2 SOLUTION Step 1. Step 2.

1 The formula holds for n ¼ 1, since a ¼ ½2a þ (1  1)d  ¼ a: 2 Assume that the formula holds for n ¼ k. Then k a þ (a þ d ) þ (a þ 2d ) þ    þ ½a þ (k  1)d  ¼ ½2a þ (k  1)d : 2

Add the (k þ 1)th term, which is (a þ kd ), to both sides of the latter equation. Then a þ (a þ d ) þ (a þ 2d ) þ    þ ½a þ (k  1)d  þ (a þ kd ) ¼

k ½2a þ (k  1)d  þ (a þ kd ): 2

k2 d kd k2 d þ kd þ 2ka þ 2a  þ a þ kd ¼ 2 2 2 kd(k þ 1) þ 2a(k þ 1) k þ 1 ¼ (2a þ kd ) ¼ 2 2

The right-hand side of this equation ¼ ka þ

which is the value of (n/2)[2a þ (n 2 1)d ] when n is replaced by (k þ 1). Hence if the formula is true for n ¼ k, we have proved it to be true for n ¼ k þ 1. But the formula holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2. Then, since it holds for n ¼ 2, it holds for n ¼ 2 þ 1 ¼ 3, and so on. Thus the formula is true for all positive integers n.

30.3

Prove by mathematical induction that, for all positive integers n, 12 þ 22 þ 32 þ    þ n2 ¼

n(n þ 1)(2n þ 1) : 6

SOLUTION Step 1.

The formula is true for n ¼ 1, since 12 ¼

1(1 þ 1)(2 þ 1) ¼ 1: 6

364

MATHEMATICAL INDUCTION

Step 2.

[CHAP. 30

Assume that the formula is true for n ¼ k: Then 12 þ 22 þ 32 þ    þ k2 ¼

k(k þ 1)(2k þ 1) : 6

Add the (k þ 1)th term, which is (k þ 1)2, to both sides of this equation. Then 12 þ 22 þ 32 þ    þ k2 þ (k þ 1)2 ¼ The right-hand side of this equation ¼ ¼

k(k þ 1)(2k þ 1) þ (k þ 1)2 : 6

k(k þ 1)(2k þ 1) þ 6(k þ 1)2 6 (k þ 1)½(2k2 þ k) þ (6k þ 6) (k þ 1)(k þ 2)(2k þ 3) ¼ 6 6

which is the value of n(n þ 1)(2n þ 1)=6 when n is replaced by (k þ 1). Hence if the formula is true for n ¼ k, it is true for n ¼ k þ 1. But the formula holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2. Then, since it holds for n ¼ 2, it holds for n ¼ 2 þ 1 ¼ 3, and so on. Thus the formula is true for all positive integers.

30.4

Prove by mathematical induction that, for all positive integers n, 1 1 1 1 n þ þ þ  þ ¼ : 13 35 57 (2n  1)(2n þ 1) 2n þ 1 SOLUTION Step 1.

The formula is true for n ¼ 1, since 1 1 1 ¼ ¼ : (2  1)(2 þ 1) 2 þ 1 3

Step 2.

Assume that the formula is true for n ¼ k. Then 1 1 1 1 k þ þ þ  þ ¼ : 13 35 57 (2k  1)(2k þ 1) 2k þ 1

Add the (k þ 1)th term, which is 1 , (2k þ 1)(2k þ 3) to both sides of the above equation. Then 1 1 1 1 1 k 1 þ þ þ  þ þ ¼ þ : 13 35 57 (2k  1)(2k þ 1) (2k þ 1)(2k þ 3) 2k þ 1 (2k þ 1)(2k þ 3) The right-hand side of this equation is k(2k þ 3) þ 1 kþ1 ¼ ; (2k þ 1)(2k þ 3) 2k þ 3 which is the value of n=(2n þ 1) when n is replaced by (k þ 1). Hence if the formula is true for n ¼ k, it is true for n ¼ k þ 1. But the formula holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2. Then, since it holds for n ¼ 2, it holds for n ¼ 2 þ 1 ¼ 3, and so on. Thus the formula is true for all positive integers n.

30.5

Prove by mathematical induction that a2n  b2n is divisible by a þ b when n is any positive integer. SOLUTION Step 1. Step 2.

The theorem is true for n ¼ 1, since a2  b2 ¼ (a þ b)(a  b): Assume that the theorem is true for n ¼ k. Then a2k  b2k is divisible by a þ b:

CHAP. 30]

MATHEMATICAL INDUCTION

365

We must show that a2kþ2  b2kþ2 is divisible by a þ b. From the identity a2kþ2  b2kþ2 ¼ a2 (a2k  b2k ) þ b2k (a2  b2 ) it follows that a2kþ2  b2kþ2 is divisible by a þ b if a2k  b2k is. Hence if the theorem is true for n ¼ k, we have proved it to be true for n ¼ k þ 1. But the theorem holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2. Then, since it holds for n ¼ 2, it holds for n ¼ 2 þ 1 ¼ 3, and so on. Thus the theorem is true for all positive integers n.

30.6

Prove the binomial formula (a þ x)n ¼ an þ nan1 x þ

n(n  1) n2 2 n(n  1)    (n  r þ 2) nrþ1 r1 a x þþ a x þ    þ xn 2! (r  1)!

for positive integers n. SOLUTION Step 1. Step 2.

The formula is true for n ¼ 1. Assume the formula is true for n ¼ k. Then

(a þ x)k ¼ ak þ nak1 x þ

k(k  1) k2 2 k(k  1)    (k  r þ 2) krþ1 r1 x þ    þ xk a x þþ a 2! ðr  1Þ!

Multiply both sides by a þ x. The multiplication on the right may be written k(k  1) k1 2 k(k  1)    (k  r þ 2) krþ2 r1 x þ    þ axk a x þ  þ a 2! (r  1)! k(k  1)    (k  r þ 3) krþ2 r1 a þ ak x þ kak1 x2 þ    þ x þ    þ xkþ1 : (r  2)! k(k  1)    (k  r þ 2) krþ2 r1 k(k  1)    (k  r þ 3) krþ2 r1 x þ x a a (r  1)! (r  2)!  k(k  1)    (k  r þ 3) krþ2 r1 k  r þ 2 a þ1 ¼ x (r  2)! r1 (k þ 1) k(k  1)    (k  r þ 3) krþ2 r1 x , a ¼ (r  1)!

akþ1 þ kak x þ

Since

the product may be written (a þ x)kþ1 ¼ akþ1 þ (k þ 1)ak x þ    þ

(k þ 1) k(k  1)    (k  r þ 3) krþ2 r1 a x þ    þ xkþ1 (r  1)!

which is the binomial formula with n replaced by k þ 1. Hence if the formula is true for n ¼ k, it is true for n ¼ k þ 1. But the formula holds for n ¼ 1; hence it holds for n ¼ 1 þ 1 ¼ 2, and so on. Thus the formula is true for all positive integers n.

30.7

Prove by mathematical induction that the sum of the interior angles, S(n), of a convex polygon is S(n) ¼ (n 2 2)1808, where n is the number of sides on the polygon. SOLUTION Step 1. Since a polygon has at least 3 sides, we start with n ¼ 3. For n ¼ 3, S(3) ¼ (3 2 2)1808 ¼ (1)1808 ¼ 1808. This is true since the sum of the interior angles of a triangle is 1808. Step 2. Assume that for n ¼ k, the formula is true. Then S(k) ¼ (k 2 2)1808 is true. Now consider a convex polygon with k þ 1 sides. We can draw in a diagonal that forms a triangle with two of the sides of the polygon. The diagonal also forms a k-sided polygon with the other sides of the original polygon. The sum of the interior

366

MATHEMATICAL INDUCTION

[CHAP. 30

angles of the (k þ l)-sided polygon, S(k þ 1), is equal to the sum of the interior angles of the triangle, S(3), and the sum of the interior angles of the k-sided polygon, S(k). S(k þ 1) ¼ S(3) þ S(k) ¼ 1808 þ (k  2)1808 ¼ ½1 þ (k  2)1808 ¼ ½(k þ 1)  21808: Hence, if the formula is true for n ¼ k, it is true for n ¼ k þ 1. Since the formula is true for n ¼ 3, and whenever it is true for n ¼ k it is true for n ¼ k þ 1, the formula is true for all positive integers n  3.

30.8

Prove by mathematical induction that n3 þ 1  n2 þ n for all positive integers. SOLUTION Step 1. For n ¼ 1, n3 þ 1 ¼ 13 þ 1 ¼ 1 þ 1 ¼ 2 and n2 þ n ¼ 12 þ 1 ¼ 1 þ 1 ¼ 2. So n3 þ 1  n2 þ n is true when n ¼ 1. Step 2.

Assume the statement is true for n ¼ k. So k3 þ 1  k2 þ k is true.

For n ¼ k þ 1, (k þ 1)3 þ 1 ¼ k3 þ 3k2 þ 3k þ 1 þ 1 ¼ k3 þ 3k2 þ 3k þ 2 ¼ k3 þ 2k2 þ k2 þ 3k þ 2 ¼ (k3 þ 2k2 ) þ (k þ 1)(k þ 2) ¼ (k3 þ 2k2 ) þ (k þ 1)½(k þ 1) þ 1 ¼ (k3 þ 2k2 ) þ ½(k þ 1)2 þ (k þ 1) We know n  1, so k  1 and k3 þ 2k2  3. Thus (k þ 1)3 þ 1  (k þ 1)2 þ (k þ 1). Hence, when the statement is true for n ¼ k, it is true for n ¼ k þ 1. Since the statement is true for n ¼ 1, and whenever it is true for n ¼ k, it is true for n ¼ k þ 1, the statement is true for all positive integers n.

Supplementary Problems Prove each of the following by mathematical induction. In each case n is a positive integer. 30.9

1 þ 3 þ 5 þ    þ (2n  1) ¼ n2

30.10

1 þ 3 þ 32 þ    þ 3n1 ¼

30.11

13 þ 23 þ 33 þ    þ n3 ¼

30.12

3n  1 2

n2 (n þ 1)2 4 a(r n  1) 2 n1 , r=1 ¼ a þ ar þ ar þ    þ ar r1

30.13

1 1 1 1 n þ þ þ  þ ¼ 12 23 34 n(n þ 1) n þ 1

30.14

1  3 þ 2  32 þ 3  33 þ    þ n  3n ¼

30.15

1 1 1 1 n þ þ þ  þ ¼ 2  5 5  8 8  11 (3n  1)(3n þ 2) 6n þ 4

30.16

1 1 1 1 n(n þ 3) þ þ þ  þ ¼ 123 234 345 n(n þ 1)(n þ 2) 4(n þ 1)(n þ 2)

30.17

an  bn is divisible by a  b, for n ¼ positive integer.

(2n  1)3nþ1 þ 3 4

CHAP. 30]

MATHEMATICAL INDUCTION

30.18

a2n1 þ b2n1 is divisible by a þ b, for n ¼ positive integer.

30.19

1  2  3 þ 2  3  4 þ    þ n(n þ 1)(n þ 2) ¼

30.20

1 þ 2 þ 22 þ    þ 2n1 ¼ 2n  1

30.21

(ab)n ¼ an bn , for n ¼ a positive integer  a n an ¼ n , for n ¼ a positive integer b b

30.22 30.23

n2 þ n is even

30.24

n3 þ 5n is divisible by 3

30.25

5n  1 is divisible by 4

30.26

4n  1 is divisible by 3

30.27

n(n þ 1)(n þ 2) is divisible by 6

30.28

n(n þ 1)(n þ 2)(n þ 3) is divisible by 24

30.29

n2 þ 1 . n

30.30

2n  n þ 1

n(n þ 1)(n þ 2)(n þ 3) 4

367

CHAPTER 31

Partial Fractions 31.1

RATIONAL FRACTIONS

A rational fraction in x is the quotient Thus

31.2

x3

P(x) of two polynomials in x. Q(x) 3x2  1 is a rational fraction: þ 7x2  4

PROPER FRACTIONS

A proper fraction is one in which the degree of the numerator is less than the degree of the denominator. Thus

x2

2x  3 þ 5x þ 4

and

4x2 þ 1 are proper fractions: x4  3x

An improper fraction is one in which the degree of the numerator is greater than or equal to the degree of the denominator. Thus

2x3 þ 6x2  9 is an improper fraction. x2  3x þ 2

By division, an improper fraction may always be written as the sum of a polynomial and a proper fraction. Thus

31.3

2x3 þ 6x2  9 32x  33 ¼ 2x þ 12 þ 2 : x2  3x þ 2 x  3x þ 2

PARTIAL FRACTIONS

A given proper fraction may often be written as the sum of other fractions (called partial fractions) whose denominators are of lower degree than the denominator of the given fraction. EXAMPLE 31.1. 3x  5 3x  5 2 1 ¼ ¼ þ : x2  3x þ 2 (x  1)(x  2) x  1 x  2

368

CHAP. 31]

31.4

PARTIAL FRACTIONS

369

IDENTICALLY EQUAL POLYNOMIALS

If two polynomials of degree n in the same variable x are equal for more than n values of x, the coefficients of like powers of x are equal and the two polynomials are identically equal. If a term is missing in either of the polynomials, it can be written in with a coefficient of 0. 31.5

FUNDAMENTAL THEOREM

A proper fraction may be written as the sum of partial fractions according to the following rules. (1) Linear factors none of which are repeated If a linear factor ax þ b occurs once as a factor of the denominator of the given fraction, then corresponding to this factor associate a partial fraction A/(ax þ b), where A is a constant = 0. EXAMPLE 31.2. xþ4 A B ¼ þ (x þ 7)(2x  1) x þ 7 2x  1

(2) Linear factors some of which are repeated If a linear factor ax þ b occurs p times as a factor of the denominator of the given fraction, then corresponding to this factor associate the p partial fractions A1 A2 Ap þ þ  þ ax þ b (ax þ b)2 (ax þ b) p where A1 , A2 , . . . , Ap are constants and Ap = 0. EXAMPLES 31.3. (a)

3x  1 A B ¼ þ (x þ 4)2 x þ 4 (x þ 4)2

(b)

5x2  2 A B C D E ¼ þ þ þ þ x3 (x þ 1)2 x3 x2 x (x þ 1)2 x þ 1

(3) Quadratic factors none of which are repeated If a quadratic factor ax2 þ bx þ c occurs once as a factor of the denominator of the given fraction, then corresponding to this factor associate a partial fraction Ax þ B ax2 þ bx þ c where A and B are constants which are not both zero. Note. It is assumed that ax2 þ bx þ c cannot be factored into two real linear factors with integer coefficients. EXAMPLES 31.4. (a)

x2  3 A Bx þ C þ ¼ (x  2)(x2 þ 4) x  2 x2 þ 4

(b)

2x3  6 A Bx þ C Dx þ E ¼ þ þ x(2x2 þ 3x þ 8)(x2 þ x þ 1) x 2x2 þ 3x þ 8 x2 þ x þ 1

370

(4)

PARTIAL FRACTIONS

[CHAP. 31

Quadratic factors some of which are repeated If a quadratic factor ax2 þ bx þ c occurs p times as a factor of the denominator of the given fraction, then corresponding to this factor associate the p partial fractions A1 x þ B 1 A2 x þ B2 Ap x þ Bp þ þ  þ 2 2 2 þ bx þ ac (ax þ bx þ c) (ax þ bx þ c) p

ax2

where A1 , B1 , A2 , B2 , . . . , Ap , Bp are constants and Ap , Bp are not both zero. EXAMPLE 31.5. (x2

31.6

x2  4x þ 1 Ax þ B Cx þ D Ex þ F ¼ 2 þ 2 þ 2 2 2 2 þ 1 þxþ1 x x (x þ 1) þ 1) (x þ x þ 1)

FINDING THE PARTIAL FRACTION DECOMPOSITION

Once the form of the partial fraction decomposition of a rational fraction has been determined, the next step is to find the system of equations to be solved to get the values of the constants needed in the partial fraction decomposition. The solution of the system of equations can be aided by the use of a graphing calculator, especially when using the matrix methods discussed in Chapter 29. Although the system of equations usually involves more than three equations, it is often quite easy to determine the value of one or two variables or relationships among the variables that allow the system to be reduced to a size small enough to be solved conveniently by any method. The methods discussed in Chapter 15 and Chapter 28 are the basic procedures used. EXAMPLE 31.6.

Find the partial fraction decomposition of

3x2 þ 3x þ 7 : (x  2)2 (x2 þ 1)

Using Rules (2) and (3) in Section 31.5, the form of the decomposition is: 3x2 þ 3x þ 7 A B Cx þ D þ 2 þ ¼ x þ1 (x  2)2 (x2 þ 1) x  2 (x  2)2 3x2 þ 3x þ 7 A(x  2)(x2 þ 1) þ B(x2 þ 1) þ (Cx þ D)(x  2)2 ¼ (x  2)2 (x2 þ 1) (x  2)2 (x2 þ 1) 3x2 þ 3x þ 7 ¼ Ax3  2Ax2 þ Ax  2A þ Bx2 þ B þ Cx3  4Cx2 þ Dx2 þ 4Cx  4Dx þ 4D 3x2 þ 3x þ 7 ¼ (A þ C)x3 þ (2A þ B 4C þ D)x2 þ (A þ 4C  4D)x þ (2A þ B þ 4D) Equating the coefficients of the corresponding terms in the two polynomials and setting the others equal to 0, we get the system of equations to solve. AþC ¼0 2A þ B  4C þ D ¼ 3 A þ 4C  4D ¼ 3 2A þ B þ 4D ¼ 7 Solving the system, we get A ¼ 1, B ¼ 5, C ¼ 1, and D ¼ 0. Thus, the partial fraction decomposition is: 3x2 þ 3x þ 7 1 5 x þ ¼ þ (x  2)2 (x2 þ 1) x  2 (x  2)2 x2 þ 1

CHAP. 31]

371

PARTIAL FRACTIONS

Solved Problems 31.1

Resolve into partial fractions 2x2

xþ2  7x  15

or

xþ2 : (2x þ 3)(x  5)

SOLUTION Let

xþ2 A B A(x  5) þ B(2x þ 3) (A þ 2B)x þ 3B  5A ¼ þ ¼ ¼ . (2x þ 3)(x  5) 2x þ 3 x  5 (2x þ 3)(x  5) (2x þ 3)(x  5)

We must find the constants A and B such that xþ2 (A þ 2B)x þ 3B  5A ¼ identically (2x þ 3)(x  5) (2x þ 3)(x  5) x þ 2 ¼ (A þ 2B)x þ 3B  5A:

or

Equating coefficients of like powers of x, we have 1 ¼ A þ 2B and 2 ¼ 3B 2 5A which when solved simultaneously give A ¼ 1=13, B ¼ 7=13. xþ2 1=13 7=13 1 7 þ ¼ þ : ¼ 2x2  7x  15 2x þ 3 x  5 13(2x þ 3) 13(x  5)

Hence Another method.

x þ 2 ¼ A(x  5) þ B(2x þ 3)

To find B, let x ¼ 5:

5 þ 2 ¼ A(0) þ B(10 þ 3), 7 ¼ 13B, B ¼ 7/13.

To find A, let x ¼ 23/2: 23/2 þ 2 ¼ A(23/2 2 5) þ B(0), 1/2 ¼ 213A/2, A ¼ 21/13.

31.2

2x2 þ 10x  3 A B C þ þ ¼ (x þ 1)(x2  9) x þ 1 x þ 3 x  3

SOLUTION 2x2 þ 10x  3 ¼ A(x2  9) þ B(x þ 1)(x  3) þ C(x þ 1)(x þ 3)

Hence

31.3

To find A, let x ¼ 2l:

2 2 10 2 3 ¼ A(1 2 9),

A ¼ 11/8.

To find B, let x ¼ 23:

18 2 30 2 3 ¼ B(23 þ 1)(23 2 3),

B ¼ 25/4.

To find C, let x ¼ 3:

18 þ 30 2 3 ¼ C(3 þ 1)(3 þ 3),

C ¼ 15/8.

2x2 þ 10x  3 11 5 15 ¼  þ : (x þ 1)(x2  9) 8(x þ 1) 4(x þ 3) 8(x  3)

2x2 þ 7x þ 23 A B C þ ¼ þ (x  1)(x þ 3)2 x  1 (x þ 3)2 x þ 3

SOLUTION 2x2 þ 7x þ 23 ¼ A(x þ 3)2 þ B(x  1) þ C(x  1)(x þ 3) ¼ A(x2 þ 6x þ 9) þ B(x  1) þ C(x2 þ 2x  3) ¼ Ax2 þ 6Ax þ 9A þ Bx  B þ Cx2 þ 2Cx  3C ¼ (A þ C)x2 þ (6A þ B þ 2C)x þ 9A  B  3C

372

PARTIAL FRACTIONS

[CHAP. 31

Equating coefficients of like powers of x, A þ C ¼ 2, 6A þ B þ 2C ¼ 7 and 9A2 B 2 3C ¼ 23. Solving simultaneously, A ¼ 2, B ¼ 25, C ¼ 0. Hence

2x2 þ 7x þ 23 2 5 ¼ :  (x  1)(x þ 3)2 x  1 (x þ 3)2

Another method. 2x2 þ 7x þ 23 ¼ A(x þ 3)2 þ B(x  1) þ C(x  1)(x þ 3)

31.4

To find A, let x ¼ 1:

2 þ 7 þ 23 ¼ A(1 þ 3)2,

A ¼ 2.

To find B, let x ¼ 23:

18 2 21 þ23 ¼ B(23 21),

B ¼ 25.

To find C, let x ¼ 0:

23 ¼ 2(3) 2 5(21) þ C(21)(3),

C ¼ 0.

2

x2  6x þ 2 A B C D ¼ 2þ þ þ 2 2 2 x x (x  2) x2 x (x  2) SOLUTION x2  6x þ 2 ¼ A(x  2)2 þ Bx(x  2)2 þ Cx2 þ Dx2 (x  2) ¼ A(x2  4x þ 4) þ Bx(x2  4x þ 4) þ Cx2 þ Dx2 (x  2) ¼ (B þ D)x3 þ (A  4B þ C  2D)x2 þ (4A þ 4B)x þ 4A Equating coefficients of like powers of x, B þ D ¼ 0, A 2 4B þ C 2 2D ¼ 1, 24A þ 4B ¼ 26, 4A ¼ 2. The simultaneous solution of these four equations is A ¼ 1/2, B ¼ 21, C ¼ 23/2, D ¼ 1. Hence

x2  6x þ 2 1 1 3 1 ¼ 2  þ 2x x 2(x  2)2 x  2 x2 (x  2)2

Another method. x2  6x þ 2 ¼ A(x  2)2 þ Bx(x  2)2 þ Cx2 þ Dx2 (x  2) To find A, let x ¼ 0: 2 ¼ 4A, A ¼ 1/2. To find C, let x ¼ 2: 4 2 12 þ 2 ¼ 4C, C ¼ 23/2. To find B and D, let x ¼ any values except 0 and 2 (for example, let x ¼ 1, x ¼ 21). Let x ¼ 1: Let x ¼ 21:

1 2 6 þ 2 ¼ A(1 2 2)2 þ B(1 2 2)2 þ C þ D(1 2 2) 1 þ 6 þ 2 ¼ A(21 2 2)2 2 B(21 2 2)2 þ C þ D(21 2 2)

and (1) B 2 D ¼ 22. and (2) 9B þ 3D ¼ 26.

The simultaneous solution of equations (1) and (2) is B ¼ 21, D ¼ 1.

31.5

x2  4x  15 . (x þ 2)3

Let y ¼ x þ 2; then x ¼ y 2 2.

SOLUTION x2  4x  15 (y  2)2  4( y  2)  15 y2  8y  3 ¼ ¼ y3 y3 (x þ 2)3 ¼

1 8 3 1 8 3   ¼   y y2 y3 x þ 2 (x þ 2)2 (x þ 2)3

CHAP. 31]

31.6

373

PARTIAL FRACTIONS

7x2  25x þ 6 Ax þ B C ¼ 2 þ 2 (x  2x  1)(3x  2) x  2x  1 3x  2 SOLUTION 7x2  25x þ 6 ¼ (Ax þ B)(3x  2) þ C(x2  2x  1) ¼ (3Ax2 þ 3Bx  2Ax  2B) þ Cx2  2Cx  C ¼ (3A þ C)x2 þ (3B  2A  2C)x þ (2B  C) Equating coefficients of like powers of x, 3A þ C ¼ 7, 3B  2A  2C ¼ 25, 2B  C ¼ 6. The simultaneous solution of these three equations is A ¼ 1, B ¼ 5, C ¼ 4: 7x2  25x þ 6 x5 4 ¼ þ : (x2  2x  1)(3x  2) x2  2x  1 3x  2

Hence

31.7

4x2  28 4x2  28 Ax þ B Cx þ D ¼ 2 ¼ þ 2 2 x 2 þ x  6 (x þ 3)(x2  2) x2 þ 3

x4

SOLUTION 4x2  28 ¼ (Ax þ B)(x2  2) þ (Cx þ D)(x2 þ 3) ¼ (Ax3 þ Bx2  2Ax  2B) þ (Cx3 þ Dx2 þ 3Cx þ 3D) ¼ (A þ C)x3 þ (B þ D)x2 þ (3C  2A)x  2B þ 3D Equating coefficients of like powers of x, A þ C ¼ 0, B þ D ¼ 4, 3C  2A ¼ 0, 2B þ 3D ¼ 28: Solving simultaneously, A ¼ 0, B ¼ 8, C ¼ 0, D ¼ 4. 4x2  28 8 4 ¼  : x4 þ x2  6 x2 þ 3 x2  2

Hence

Supplementary Problems Find the partial fraction decomposition of each rational fraction. 31.8 31.9 31.10 31.11 31.12

x2

xþ2  7x þ 12

12x þ 11 x2 þ x  6 2x2

8x þ 3x  2

5x þ 4 x2 þ 2x x2

x  3x  18

31.13

10x2 þ 9x  7 (x þ 2)(x2  1)

31.18

31.14

x2  9x  6 x3 þ x2  6x

31.19

31.15

x2

x3 4

31.20

5x2 þ 8x þ 21 þ x þ 6)(x þ 1)

(x2

5x3 þ 4x2 þ 7x þ 3 (x2 þ 2x þ 2)(x2  x  1) x3

3x 1

31.16

3x2  8x þ 9 (x  2)3

31.21

7x3 þ 16x2 þ 20x þ 5 (x2 þ 2x þ 2)2

31.17

3x3 þ 10x2 þ 27x þ 27 x2 (x þ 3)2

31.22

7x  9 (x þ 1)(x  3)

374

PARTIAL FRACTIONS

[CHAP. 31

31.23

x þ 10 x(x  2)(x þ 2)

31.26

5x2 þ 3x þ 1 (x þ 2)(x2 þ 1)

31.29

x3 (x2 þ 4)2

31.24

3x  1 x2  1

31.27

2x þ 9 (2x þ 1)(4x2 þ 9)

31.30

x4 þ 3x2 þ x þ 1 (x þ 1)(x2 þ 1)2

31.25

x3

7x  2  x2  2x

31.28

2x3  x þ 3 þ 4)(x2 þ 1)

(x2

ANSWERS TO SUPPLEMENTARY PROBLEMS 31.8

6 5  x4 x3

31.16

3 4 5 þ þ x  2 (x  2)2 (x  2)3

31.24

1 2 þ x1 xþ1

31.9

7 5 þ x2 xþ3

31.17

1 3 2 5 þ þ  x x2 x þ 3 (x þ 3)2

31.25

1 2 3 þ þ x x2 xþ1

31.10

3 2  2x  1 x þ 2

31.18

2x þ 3 3 þ x2 þ x þ 6 x þ 1

31.26

3 2x  1 þ x þ 2 x2 þ 1

31.11

2 3 þ x xþ2

31.19

2x  1 3x þ 1 þ x2 þ 2x þ 2 x2  x  1

31.27

1 2x þ 2x þ 1 4x2 þ 9

31.12

2=3 1=3 þ x6 xþ3

31.20

1 x þ 1 þ x  1 x2 þ x þ 1

31.28

3x  1 x þ 1 þ x2 þ 4 x2 þ 1

31.13

3 2 5 þ þ xþ1 x1 xþ2

31.21

7x þ 2 2x þ 1 þ x2 þ 2x þ 2 (x2 þ 2x þ 2)2

31.29

x 4x þ x2 þ 4 (x2 þ 4)2

31.14

1 2 2  þ x x2 xþ3

31.22

4 3 þ xþ1 x3

31.30

1 x þ x þ 1 (x2 þ 1)2

31.15

xþ

2 2 þ x2 xþ2

31.23

5=2 3=2 1 þ þ x x2 xþ2

APPENDIX A

Table of Common Logarithms N

0

1

2

3

4

5

6

7

8

9

10 11 12 13 14

0000 0414 0792 1139 1461

0043 0453 0828 1173 1492

0086 0492 0864 1206 1523

0128 0531 0899 1239 1553

0170 0569 0934 1271 1584

0212 0607 0969 1303 1614

0253 0645 1004 1335 1644

0294 0682 1038 1367 1673

0334 0719 1072 1399 1703

0374 0755 1106 1430 1732

15 16 17 18 19

1761 2041 2304 2553 2788

1790 2068 2330 2577 2810

1818 2095 2355 2601 2833

1847 2122 2380 2625 2856

1875 2148 2405 2648 2878

1903 2175 2430 2672 2900

1931 2201 2455 2695 2923

1959 2227 2480 2718 2945

1987 2253 2504 2742 2967

2014 2279 2529 2765 2989

20 21 22 23 24

3010 3222 3424 3617 3802

3032 3243 3444 3636 3820

3054 3263 3464 3655 3838

3075 3284 3483 3674 3856

3096 3304 3502 3692 3874

3118 3324 3522 3711 3892

3139 3345 3541 3729 3909

3160 3365 3560 3747 3927

3181 3385 3579 3766 3945

3201 3404 3598 3784 3962

25 26 27 28 29

3979 4150 4314 4472 4624

3997 4166 4330 4487 4639

4014 4183 4346 4502 4654

4031 4200 4362 4518 4669

4048 4216 4378 4533 4683

4065 4232 4393 4548 4698

4082 4249 4409 4564 4713

4099 4265 4425 4579 4728

4116 4281 4440 4594 4742

4133 4298 4456 4609 4757

30 31 32 33 34

4771 4914 5051 5185 5315

4786 4928 5065 5198 5328

4800 4942 5079 5211 5340

4814 4955 5092 5224 5353

4829 4969 5105 5237 5366

4843 4983 5119 5250 5378

4857 4997 5132 5263 5391

4871 5011 5145 5276 5403

4886 5024 5159 5289 5416

4900 5038 5172 5302 5428

N

0

1

2

3

4

5

6

7

8

9

375

376

TABLE OF COMMON LOGARITHMS

[APP. A

N

0

1

2

3

4

5

6

7

8

9

35 36 37 38 39

5441 5563 5682 5798 5911

5453 5575 5694 5809 5922

5465 5587 5705 5821 5933

5478 5599 5717 5832 5944

5490 5611 5729 5843 5955

5502 5623 5740 5855 5966

5514 5635 5752 5866 5977

5527 5647 5763 5877 5988

5539 5658 5775 5888 5999

5551 5670 5786 5899 6010

40 41 42 43 44

6021 6128 6232 6335 6435

6031 6138 6243 6345 6444

6042 6149 6253 6355 6454

6053 6160 6263 6365 6464

6064 6170 6274 6375 6474

6075 6180 6284 6385 6484

6085 6191 6294 6395 6493

6096 6201 6304 6405 6503

6107 6212 6314 6415 6513

6117 6222 6325 6425 6522

45 46 47 48 49

6532 6628 6721 6812 6902

6542 6637 6730 6821 6911

6551 6646 6739 6830 6920

6561 6656 6749 6839 6928

6571 6665 6758 6848 6937

6580 6675 6767 6857 6946

6590 6684 6776 6866 6955

6599 6693 6785 6875 6964

6609 6702 6794 6884 6972

6618 6712 6803 6893 6981

50 51 52 53 54

6990 7076 7160 7243 7324

6998 7084 7168 7251 7332

7007 7093 7177 7259 7340

7016 7101 7185 7267 7348

7024 7110 7193 7275 7356

7033 7118 7202 7284 7364

7042 7126 7210 7292 7372

7050 7135 7218 7300 7380

7059 7143 7226 7308 7388

7067 7152 7235 7316 7396

55 56 57 58 59

7404 7482 7559 7634 7709

7412 7490 7566 7642 7716

7419 7497 7574 7649 7723

7427 7505 7582 7657 7731

7435 7513 7589 7664 7738

7443 7520 7597 7672 7745

7451 7528 7604 7679 7752

7459 7536 7612 7686 7760

7466 7543 7619 7694 7767

7474 7551 7627 7701 7774

60 61 62 63 64

7782 7853 7924 7993 8062

7789 7860 7931 8000 8069

7796 7868 7938 8007 8075

7803 7875 7945 8014 8082

7810 7882 7952 8021 8089

7818 7889 7959 8028 8096

7825 7896 7966 8035 8102

7832 7903 7973 8041 8109

7839 7910 7980 8048 8116

7846 7917 7987 8055 8122

65 66 67 68 69

8129 8195 8261 8325 8388

8136 8202 8267 8331 8395

8142 8209 8274 8338 8401

8149 8215 8280 8344 8407

8156 8222 8287 8351 8414

8162 8228 8293 8357 8420

8169 8235 8299 8363 8426

8176 8241 8306 8370 8432

8182 8248 8312 8376 8439

8189 8254 8319 8382 8445

70 71 72 73 74

8451 8513 8573 8633 8692

8457 8519 8579 8639 8698

8463 8525 8585 8645 8704

8470 8531 8591 8651 8710

8476 8537 8597 8657 8716

8482 8543 8603 8663 8722

8488 8549 8609 8669 8727

8494 8555 8615 8675 8733

8500 8561 8621 8681 8739

8506 8567 8627 8686 8745

75 76 77 78 79

8751 8808 8865 8921 8976

8756 8814 8871 8927 8982

8762 8820 8876 8932 8987

8768 8825 8882 8938 8993

8774 8831 8887 8943 8998

8779 8837 8893 8949 9004

8785 8842 8899 8954 9009

8791 8848 8904 8960 9015

8797 8854 8910 8965 9020

8802 8859 8915 8971 9025

N

0

1

2

3

4

5

6

7

8

9

APP. A]

377

TABLE OF COMMON LOGARITHMS

N

0

1

2

3

4

5

6

7

8

9

80 81 82 83 84

9031 9085 9138 9191 9243

9036 9090 9143 9196 9248

9042 9096 9149 9201 9253

9047 9101 9154 9206 9258

9053 9106 9159 9212 9263

9058 9112 9165 9217 9269

9063 9117 9170 9222 9274

9069 9122 9175 9227 9279

9074 9128 9180 9232 9284

9079 9133 9186 9238 9289

85 86 87 88 89

9294 9345 9395 9445 9494

9299 9350 9400 9450 9499

9304 9355 9405 9455 9504

9309 9360 9410 9460 9509

9315 9365 9415 9465 9513

9320 9370 9420 9469 9518

9325 9375 9425 9474 9523

9330 9380 9430 9479 9528

9335 9385 9435 9484 9533

9340 9390 9440 9489 9538

90 91 92 93 94

9542 9590 9638 9685 9731

9547 9595 9643 9689 9736

9552 9600 9647 9694 9741

9557 9605 9652 9699 9745

9562 9609 9657 9703 9750

9566 9614 9661 9708 9754

9571 9619 9666 9713 9759

9576 9624 9671 9717 9763

9581 9628 9675 9722 9768

9586 9633 9680 9727 9773

95 96 97 98 99

9777 9823 9868 9912 9956

9782 9827 9872 9917 9961

9786 9832 9877 9921 9965

9791 9836 9881 9926 9969

9795 9841 9886 9930 9974

9800 9845 9890 9934 9978

9805 9850 9894 9939 9983

9809 9854 9899 9943 9987

9814 9859 9903 9948 9991

9818 9863 9908 9952 9996

N

0

1

2

3

4

5

6

7

8

9

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APPENDIX B

Table of Natural Logarithms N

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

1.0 1.1 1.2 1.3 1.4

0.0000 0.0953 0.1823 0.2624 0.3365

0.0100 0.1044 0.1906 0.2700 0.3436

0.0198 0.1133 0.1989 0.2776 0.3507

0.0296 0.1222 0.2070 0.2852 0.3577

0.0392 0.1310 0.2151 0.2927 0.3646

0.0488 0.1398 0.2231 0.3001 0.3716

0.0583 0.1484 0.2311 0.3075 0.3784

0.0677 0.1570 0.2390 0.3148 0.3853

0.0770 0.1655 0.2469 0.3221 0.3920

0.0862 0.1740 0.2546 0.3293 0.3988

1.5 1.6 1.7 1.8 1.9

0.4055 0.4700 0.5306 0.5878 0.6419

0.4121 0.4762 0.5365 0.5933 0.6471

0.4187 0.4824 0.5423 0.5988 0.6523

0.4253 0.4886 0.5481 0.6043 0.6575

0.4318 0.4947 0.5539 0.6098 0.6627

0.4383 0.5008 0.5596 0.6152 0.6678

0.4447 0.5068 0.5653 0.6206 0.6729

0.4511 0.5128 0.5710 0.6259 0.6780

0.4574 0.5188 0.5766 0.6313 0.6831

0.4637 0.5247 0.5822 0.6366 0.6881

2.0 2.1 2.2 2.3 2.4

0.6931 0.7419 0.7885 0.8329 0.8755

0.6981 0.7467 0.7930 0.8372 0.8796

0.7031 0.7514 0.7975 0.8416 0.8838

0.7080 0.7561 0.8020 0.8459 0.8879

0.7130 0.7608 0.8065 0.8502 0.8920

0.7178 0.7655 0.8109 0.8544 0.8961

0.7227 0.7701 0.8154 0.8587 0.9002

0.7275 0.7747 0.8198 0.8629 0.9042

0.7324 0.7793 0.8242 0.8671 0.9083

0.7372 0.7839 0.8286 0.8713 0.9123

2.5 2.6 2.7 2.8 2.9

0.9163 0.9555 0.9933 1.0296 1.0647

0.9203 0.9594 0.9969 1.0332 1.0682

0.9243 0.9632 1.0006 1.0367 1.0716

0.9282 0.9670 1.0043 1.0403 1.0750

0.9322 0.9708 1.0080 1.0438 1.0784

0.9361 0.9746 1.0116 1.0473 1.0818

0.9400 0.9783 1.0152 1.0508 1.0852

0.9439 0.9821 1.0188 1.0543 1.0886

0.9478 0.9858 1.0225 1.0578 1.0919

0.9517 0.9895 1.0260 1.0613 1.0953

3.0 3.1 3.2 3.3 3.4

1.0986 1.1314 1.1632 1.1939 1.2238

1.1019 1.1346 1.1663 1.1970 1.2267

1.1053 1.1378 1.1694 1.2000 1.2296

1.1086 1.1410 1.1725 1.2030 1.2326

1.1119 1.1442 1.1756 1.2060 1.2355

1.1151 1.1474 1.1787 1.2090 1.2384

1.1184 1.1506 1.1817 1.2119 1.2413

1.1217 1.1537 1.1848 1.2149 1.2442

1.1249 1.1569 1.1878 1.2179 1.2470

1.1282 1.1600 1.1909 1.2208 1.2499

N

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

379

380

TABLE OF NATURAL LOGARITHMS

[APP. B

N

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

3.5 3.6 3.7 3.8 3.9

1.2528 1.2809 1.3083 1.3350 1.3610

1.2556 1.2837 1.3110 1.3376 1.3635

1.2585 1.2865 1.3137 1.3403 1.3661

1.2613 1.2892 1.3164 1.3429 1.3686

1.2641 1.2920 1.3191 1.3455 1.3712

1.2669 1.2947 1.3218 1.3481 1.3737

1.2698 1.2975 1.3244 1.3507 1.3762

1.2726 1.3002 1.3271 1.3533 1.3788

1.2754 1.3029 1.3297 1.3558 1.3813

1.2782 1.3056 1.3324 1.3584 1.3838

4.0 4.1 4.2 4.3 4.4

1.3863 1.4110 1.4351 1.4586 1.4816

1.3888 1.4134 1.4375 1.4609 1.4839

1.3913 1.4159 1.4398 1.4633 1.4861

1.3938 1.4183 1.4422 1.4656 1.4884

1.3962 1.4207 1.4446 1.4679 1.4907

1.3987 1.4231 1.4469 1.4702 1.4929

1.4012 1.4255 1.4493 1.4725 1.4952

1.4036 1.4279 1.4516 1.4748 1.4974

1.4061 1.4303 1.4540 1.4770 1.4996

1.4085 1.4327 1.4563 1.4793 1.5019

4.5 4.6 4.7 4.8 4.9

1.5041 1.5261 1.5476 1.5686 1.5892

1.5063 1.5282 1.5497 1.5707 1.5913

1.5085 1.5304 1.5518 1.5728 1.5933

1.5107 1.5326 1.5539 1.5748 1.5953

1.5129 1.5347 1.5560 1.5769 1.5974

1.5151 1.5369 1.5581 1.5790 1.5994

1.5173 1.5390 1.5602 1.5810 1.6014

1.5195 1.5412 1.5623 1.5831 1.6034

1.5217 1.5433 1.5644 1.5851 1.6054

1.5239 1.5454 1.5665 1.5872 1.6074

5.0 5.1 5.2 5.3 5.4

1.6094 1.6292 1.6487 1.6677 1.6864

1.6114 1.6312 1.6506 1.6696 1.6882

1.6134 1.6332 1.6525 1.6715 1.6901

1.6154 1.6351 1.6544 1.6734 1.6919

1.6174 1.6371 1.6563 1.6752 1.6938

1.6194 1.6390 1.6582 1.6771 1.6956

1.6214 1.6409 1.6601 1.6790 1.6974

1.6233 1.6429 1.6620 1.6808 1.6993

1.6253 1.6448 1.6639 1.6827 1.7011

1.6273 1.6467 1.6658 1.6845 1.7029

5.5 5.6 5.7 5.8 5.9

1.7047 1.7228 1.7405 1.7579 1.7750

1.7066 1.7246 1.7422 1.7596 1.7766

1.7084 1.7263 1.7440 1.7613 1.7783

1.7102 1.7281 1.7457 1.7630 1.7800

1.7120 1.7299 1.7475 1.7647 1.7817

1.7138 1.7317 1.7492 1.7664 1.7834

1.7156 1.7334 1.7509 1.7682 1.7851

1.7174 1.7352 1.7527 1.7699 1.7867

1.7192 1.7370 1.7544 1.7716 1.7884

1.7210 1.7387 1.7561 1.7733 1.7901

6.0 6.1 6.2 6.3 6.4

1.7918 1.8083 1.8245 1.8406 1.8563

1.7934 1.8099 1.8262 1.8421 1.8579

1.7951 1.8116 1.8278 1.8437 1.8594

1.7967 1.8132 1.8294 1.8453 1.8610

1.7984 1.8148 1.8310 1.8469 1.8625

1.8001 1.8165 1.8326 1.8485 1.8641

1.8017 1.8181 1.8342 1.8500 1.8656

1.8034 1.8197 1.8358 1.8516 1.8672

1.8050 1.8213 1.8374 1.8532 1.8687

1.8066 1.8229 1.8390 1.8547 1.8703

6.5 6.6 6.7 6.8 6.9

1.8718 1.8871 1.9021 1.9169 1.9315

1.8733 1.8886 1.9036 1.9184 1.9330

1.8749 1.8901 1.9051 1.9199 1.9344

1.8764 1.8916 1.9066 1.9213 1.9359

1.8779 1.8931 1.9081 1.9228 1.9373

1.8795 1.8946 1.9095 1.9242 1.9387

1.8810 1.8961 1.9110 1.9257 1.9402

1.8825 1.8976 1.9125 1.9272 1.9416

1.8840 1.8991 1.9140 1.9286 1.9430

1.8856 1.9006 1.9155 1.9301 1.9445

7.0 7.1 7.2 7.3 7.4

1.9459 1.9601 1.9741 1.9879 2.0015

1.9473 1.9615 1.9755 1.9892 2.0028

1.9488 1.9629 1.9769 1.9906 2.0042

1.9502 1.9643 1.9782 1.9920 2.0055

1.9516 1.9657 1.9796 1.9933 2.0069

1.9530 1.9671 1.9810 1.9947 2.0082

1.9544 1.9685 1.9824 1.9961 2.0096

1.9559 1.9699 1.9838 1.9974 2.0109

1.9573 1.9713 1.9851 1.9988 2.0122

1.9587 1.9727 1.9865 2.0001 2.0136

7.5 7.6 7.7 7.8 7.9

2.0149 2.0282 2.0412 2.0541 2.0669

2.0162 2.0295 2.0425 2.0554 2.0681

2.0176 2.0308 2.0438 2.0567 2.0694

2.0189 2.0321 2.0451 2.0580 2.0707

2.0202 2.0334 2.0464 2.0592 2.0719

2.0215 2.0347 2.0477 2.0605 2.0732

2.0229 2.0360 2.0490 2.0618 2.0744

2.0242 2.0373 2.0503 2.0631 2.0757

2.0255 2.0386 2.0516 2.0643 2.0769

2.0268 2.0399 2.0528 2.0665 2.0782

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INDEX

Abscissa, 91 Absolute value, 2 Absolute value inequality, 199 Addition, 1 associative property for, 3 commutative property for, 3 of algebraic expressions, 12 of complex numbers, 68 of fractions, 4 of radicals, 59 rules of signs for, 3 Algebra fundamental operations of, 1 fundamental theorem of, 215 Algebraic expressions, 12 Antilogarithm, 265, 266 Arithmetic mean, 247 Arithmetic means, 247 Arithmetic sequence, 248 Associative properties, 3 Asymptotes, 235 horizontal, 235 vertical, 235 Augmented matrix, 354 Axioms of equality, 73, 74

Base of logarithms, 263 Base of powers, 4 Best buy, 82 Binomial, 12 Binomial coefficients, 304 Binomial expansion, 303 formula or theorem, 303 proof of, for positive integral powers, 365 Bounds, lower and upper, for roots, 217 Braces, 13 Brackets, 13

Cancellation, 41 Characteristic of a logarithm, 264

Circle, 170 Circular permutations, 289 Closure property, 22 Coefficients, 12 in binomial formula, 304 lead, 214 relation between roots and, 152 Cofactor, 328 Combinations, 289 Common difference, 245 Common logarithms, 264 Common ratio, 245 Commutative properties, 3 Completeness property, 23 Completing the square, 151 Complex fractions, 43 Complex numbers, 67 algebraic operations with, 68 conjugate of, 67 equal, 67 graphical addition and subtraction of, 69–70 imaginary part of, 67 pure imaginary, 67 real part of, 67 Compound interest, 277 Conditional equation, 73 Conditional inequality, 199 Conic sections, 169–180 circle, 170 ellipse, 172 hyperbola, 177 parabola, 171 Conjugate complex numbers, 67 Conjugate irrational numbers, 60 Consistent equations, 138 Constant, 89 of proportionality or variation, 82 Continuity, 216 Coordinate system, rectangular, 90 Cramer’s Rule, 323–326 Cube of a binomial, 27 Cubic equation, 75

383

384

Decimal, repeating, 255 Defective equations, 74 Degree, 13 of a monomial, 13 of a polynomial, 13 Denominator, 1, 41 Density property, 23 Dependent equations, 138 Dependent events, 310 Dependent variable, 91 Depressed equations, 221 Descartes’ Rules of Signs, 217–218 Determinants, 323 expansion or value of, 323, 324, 328 of order n, 326 of second order, 323 of third order, 324 properties of, 327 solution of linear equations by, 323, 325, 328–329 Difference, 1 common, 245 of two cubes, 33 of two squares, 33 tabular, 264 Discriminant, 151 Distributive law for multiplication, 3 Dividend, 1 Division, 1 by zero, 1 of algebraic expressions, 15 of complex numbers, 69 of fractions, 1, 4–5 of radicals, 60 synthetic, 215 Divisor, 1 Double root, 151, 216 Domain, 89–90 e, base of natural logarithms, 265 Effective rate of interest, 281 Element of a determinant, 323 Elementary row operations, 351 Ellipse, 173–174 Equations, 73 complex roots of, 216 conditional, 73 cubic, 75 defective, 74 degree of, 13 depressed, 221 equivalent, 74 graphs of, (see Graphs) identity, 73 irrational roots of, 152 limits for roots of, 217 linear, 75 literal, 114

INDEX

number of roots of, 216 quadratic, 75, 150 quadratic type, 153 quartic, 75 quintic, 75 radical, 152 redundant, 74 roots of, 73 simultaneous, 137 solutions of, 73 systems of, 137 transformation of, 73–74 with given roots, 222 Equivalent equations, 74 Equivalent fractions, 41 Expectations, mathematical, 311 Exponential equations, 274 Exponential form, 263 Exponents, 4, 48 applications, 280 fractional, 49 laws of, 4, 49–50 zero, 49 Extraneous roots, solutions, 74 Extremes, 81 Factor, 32 greatest common, 34 prime, 32 Factor theorem, 215 Factorial notation, 288 Factoring, 32 Failure, probability of, 310 Formulas, 74 Fourth proportional, 81 Fractional exponents, 49 Fractions, 4–5, 42–43 complex, 43 equivalent, 41 improper, 368 operations with, 4–5 partial, 368 proper, 368 rational algebraic, 41 reduced to lowest terms, 41 signs of, 4 Function, 89 graph of, 90–95 linear, 75, 128–131 notation for, 90 polynomial, 214 quadratic, 75, 150–152 Fundamental Theorem of Algebra, 215 Fundamental Counting Principle, 288 General or nth term, 245 Geometric mean, 247

INDEX

Geometric sequence, 245–246 infinite, 246 Geometric series, infinite, 246 Graphical solution of equations, 138, 191 Graphs, 90–95 of equations, 90–95, 138, 167–178 of functions, 90–95 of linear equations in two variables, 138 of quadratic equations in two variables, 191 with holes, 235 Greater than, 2 Greatest common factor, 33 Grouping, symbols of, 13 Grouping of terms, factoring by, 32 Harmonic mean, 247 Harmonic sequence, 246 Holes, in graph, 235 Homogenous linear equations, 329 Hyperbola, 177 i, 67 Identically equal polynomials, 369 Identity, 73 property, 22 matrix, 351 Imaginary numbers, 2, 67 Imaginary part of a complex number, 67 Imaginary unit, 2, 67 Improper fraction, 368 Inconsistent equations, 138 Independent events, 310 Independent variable, 89 Index, 48, 58 Index of a radical, 58 reduction of, 59 Induction, mathematical, 362 Inequalities, 199 absolute, 199 conditional, 199 graphical solution of, 202 higher order, 200 principles of, 199 sense of, 199 signs of, 199 Infinite geometric sequence or series, 246 Infinity, 246 Integers, 22 Integral roots theorem, 216 Interest, 276–277 compound, 277 simple, 276 Intermediate Value Theorem, 216 Interpolation in logarithms, 265 Interpolation, linear, 265 Inverse property, 22

Irrationality, proofs of, 78–79, 225 Irrational number, 2 Irrational roots, 152 approximating, 218 Inverse matrix, 352 Inverse property, 22 Least common denominator, 42 Least common multiple, 34 Less than, 2 Like terms, 13 Linear equations, 114 consistent, 138 dependent, 138 determinant solution of system of, 323–329 graphical solution of systems of, 138 homogenous, 329 inconsistent, 138 in one variable, 114 simultaneous, systems of, 137 Linear function, 113 Linear interpolation, 265 Linear programming, 203 Lines, 128 intercept form, 131 slope–intercept form, 130 slope–point form, 130 two-point form, 130 Literals, 13 Logarithms, 263 applications of, 277–279 base of, 263 common system of, 264 characteristic of common, 264 laws of, 263 mantissa of common, 264 natural base of, 265 natural system of, 265 tables of common, 375 tables of natural, 378 Lower bound or limit for roots, 217 Mantissa, 264 Mathematical expectation, 311 Mathematical induction, 362 Matrix, 349 addition, 349 identity, 351 inverse, 352 multiplication, 350 scalar multiplication, 350 Maximum point, relative, 101 applications, 103–106 Mean proportional, 81 Means of a proportion, 81 Minimum point, relative, 101 applications, 103–106

385

386

Minors, 328 Minuend, 14 Monomial, 12 Monomial factor, 33 Multinomial, 12 Multiplication, 1, 14–15 associative property for, 3 by zero, 2 commutative property for, 3 distributive property for, 3 of algebraic expressions, 12 of complex numbers, 68 of fractions, 4 of radicals, 60 rules of signs for, 3 Mutually exclusive events, 311 Natural logarithms, 265 Natural numbers, 2, 22 Negative numbers, 1 Nonreal roots of an equations, 152 Numbers, 2 absolute value of, 2 complex, 67 counting, 22 graphical representation of real, 2 imaginary, 67 integers, 22 irrational, 22 literal, 13 natural, 2, 22 negative, 2 operations with real, 1–5 positive, 2 prime, 22 rational, 22 real, 22 whole, 22 Number system, real, 2 Numerator, 1, 41 Numerical coefficient, 12 Odds, 311 Operations, fundamental, 1 Order of a determinant, 323, 324, 326 Order of real numbers, 2, 23 Order property, 23 Ordinate, 91 Origin, 2 of a rectangular coordinate system, 90 Parabola, 171–172 vertex of, 100 Parentheses, 13 Partial fractions, 368 Pascal’s triangle, 305

INDEX

Perfect square trinomial, 33 Perfect nth powers, 59 Permutations, 288 Point, coordinates of a, 91 Polynomials, 12 degree of, 75 factors of, 32–35 identically equal, 369 operations with, 13–15 prime, 32 relatively prime, 34 Polynomial functions, 214–234 solving, 216 zeros, 214–215 Positive numbers, 1 Powers, 4, 48 logarithms of, 263 of binomials, 27, 303–309 Prime factor, 32 number, 32 polynomial, 32 Principal, 276 Principal root, 48 Probability, 310 binomial, 311 conditional, 311–312 of dependent events, 310 of independent events, 311 of mutually exclusive events, 311 Product, 1, 4, 14 of roots of quadratic equation, 150 Products, special, 27 Proper fraction, 368 Proportion, 81 Proportional, 81 fourth, 81 mean, 81 third, 81 Proportionality, constant of, 82 Pure imaginary number, 67 Quadrants, 91 Quadratic equations, 150 discriminant of, 151 formation of, from given roots, 152 in one variable, 149 in two variables, 169 nature of roots of, 152 product of roots of, 152 simultaneous, 191–193 sum of roots of, 152 Quadratic equations in one variable, 150–153 solutions of, 150–152 by completing the square, 151 by factoring, 150–151

INDEX

by formula, 151–152 by graphical methods, 152 by square root method, 150 Quadratic equations in two variables, 169 circle, 170 discriminant, 169 ellipse, 173 hyperbola, 177 parabola, 171 Quadratic formula, proof of, 154–155 Quadratic type equations, 153 Quartic equation, 75 Quintic equation, 75 Quotient, 1, 4, 15 Radical equations, 152–153 Radicals, 58 algebraic addition of, 59 changing the form of, 58–59 equations involving, 152–153 index or order of, 58 multiplication and division of, 60–61 rationalization of denominator of, 60–61 reduction of index of, 59 removal of perfect nth powers of, 59 similar, 59 simplest form of, 59 Radicand, 58 Range of a function, 89 Rate of interest, 276 Ratio, 81 common, 245 Rational function, 235 graphing, 236–237 Rational number, 1, 22 Rational root theorem, 216 Rationalization of denominator, 60 Real numbers, 1, 22 graphical representation of, 2 Real part of a complex number, 67 Reciprocal, 4 Rectangular coordinates, 90 Redundant equations, 74 Relation, 89 Remainder, 15, 214 Remainder theorem, 214 Repeating decimal, 255 Roots, 48, 73, 210 nature of, for quadratic equation, 152 double, 151, 216 extraneous, 74 integral, 216 irrational, 152 nth, 58 number of, 216

of an equation, 73 of quadratic equations, 150 principal nth, 58 rational, 216 Row-echelon form, 352 Row equivalent matrices, 352 Scaling, 93 Scientific notation, 50 Sense of an inequality, 199 Sequence, 245 arithmetic, 245 geometric, 245–246 harmonic, 246 infinite, 246 nth or general term of a, 245, 246 Series, 245 infinite geometric, 246 Shifts, 92 horizontal, 93 vertical, 92 Signs, 3 Descartes’ Rule of, 217 in a fraction, 4 rules of, 3 Simple interest, 276 Simultaneous equations, 137, 191 Simultaneous linear equations, 137 Simultaneous quadratic equations, 191 Slope, 128 horizontal lines, 128 parallel lines, 129 perpendicular lines, 129 vertical lines, 129 Solutions, 73 extraneous, 74 graphical, 138, 191 of systems of equations, 323, 329 trivial, non-trivial, 329 Special products, 27 Square, 48 of a binomial, 27 of a trinomial, 27 Straight line, 128–131 Subtraction, 1, 4, 14 of algebraic expressions, 12 of complex numbers, 68 of fractions, 4, 42 of radicals, 59 Subtrahend, 14 Sum, 1 of arithmetic sequence, 245 of geometric sequence, 245–246 of infinite geometric sequence, 246 of roots of a quadratic equation, 150 of two cubes, 33

387

388

Symbols of grouping, 13 Symmetry, 91 Synthetic division, 215 Systems of equations, 137, 191 Systems of inequalities, 199 Systems of m equations in n unknowns, 329 Tables, 375, 378 of common logarithms, 375 of natural logarithms, 378 Tabular difference, 264 Term, 12 degree of, 13 integral and rational, 13 like or similar, 13 of sequences, 245 of series, 245 Trinomial, 12 factors of a, 33

INDEX

square of a, 27 Trivial solutions, 329 Unique Factorization Theorem, 32 Unit price, 82 Variable, 89 dependent, 90 independent, 89 Variation, 81–82 direct, 82 inverse, 82 joint, 82 Zero, 1 degree, 13 division by, 1 exponent, 49 multiplication by, 1 Zeros, 73, 214

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